MSc FM, MTF, QF Induction Programme

Advanced

Academic year 2010/2011

Lecture notes prepared by Laura Ballotta and George Harrison

Section I

Linear Algebra Differential Calculus Difference and Differential

Table Content 1 LINEAR ALGEBRA ...... 5

1.1. WHAT IS A MATRIX?...... 5 1.2. SPECIAL TYPES OF MATRICES ...... 6 1.2.1. Vectors...... 6 1.2.2. Square matrix...... 6 1.2.3. Diagonal matrix ...... 7 1.2.4. Identity Matrix...... 7 1.3. MATRIX OPERATIONS ...... 7 1.3.1 Adding and subtracting matrices ...... 7 1.3.2 Scalar multiplication...... 8 1.3.3. Transpose of a Matrix...... 8 1.3.4. Vector multiplication ...... 9 1.3.5 Matrix multiplication ...... 10 1.3.6 Properties of matrix multiplication ...... 12 1.4. THE DETERMINANT...... 13 1.4.1. How to find the determinant of a (2 x 2) matrix ...... 13 1.4.2. The determinant of a (3 x 3) matrix ...... 14 1.4.3. Method of cofactors in determining determinants ...... 14 1.4.4. Properties of Determinants ...... 17 1.5. INVERSE OF A MATRIX...... 18 1.5.1. Definition of an inverse matrix...... 18 1.5.2. How to find the inverse of a square matrix...... 18 1.6. TRACE OF A MATRIX...... 19 1.7. SYSTEM OF LINEAR EQUATIONS ...... 20 1.8. PARTITIONED MATRICES...... 22 1.9. VECTOR SPACES ...... 22 1.9.1 Rank of a matrix ...... 25 1.10 EIGENVALUES AND EIGENVECTORS ...... 26 1.10.1 General results for characteristic roots and vectors ...... 28 1.10.2 Further results on determining the rank of a matrix...... 29 1.11 DEFINITE MATRICES ...... 29 2. ...... 32

2.1. DIFFERENTIATION ...... 32 2.1.1. Definitions ...... 32 2.1.2. Rules of Differentiation ...... 35 2.1.3. of Exponential and logarithmic functions...... 37 2.1.4. Higher order derivatives...... 38 2.2. APPLICATIONS OF DERIVATIVES ...... 39 2.2.1. Taylor and Maclaurin ...... 39 2.2.2. Increasing and decreasing functions ...... 41 2.2.3. Concavity and Convexity...... 42 2.2.4. First test for relative extremum...... 44 2.2.5. Second for relative extremum...... 45 2.2.6. The Nth-derivative test...... 46 2.2.7. Economic applications of derivatives ...... 46 3. DIFFERENTIAL CALCULUS (THE MULTIVARIATE CASE) ...... 53

3.1. MULTIVARIATE FUNCTIONS ...... 53 3.2 LIMITS AND CONTINUITY...... 53 3.3 DIFFERENTIATION ...... 55 3.4 APPLICATION OF DIFFERENTIATION ...... 61 3.4.1 Critical Values of Bivariate Functions...... 61 3.4.2. Critical Points of Multivariate Functions...... 64 3.4.3. Constrained Optimisation ...... 67 4 INTEGRAL CALCULUS...... 72

4.1. INTEGRATION...... 72

2 4.1.1. Definite and Indefinite ...... 72 4.1.2. Properties of definite integrals...... 74 4.1.3. Rules of Integration - Indefinite Integral ...... 75 4.1.4. Auxiliary Conditions...... 79 4.2. APPLICATIONS OF INTEGRALS ...... 79 4.2.1. Deriving Totals (Revenue/Cost) from Marginals (Revenue/Cost)...... 79 4.2.2. Present Value of Cashflows...... 80 4.2.3. Measuring Probabilities ...... 81 4.3 MULTIPLE INTEGRATION...... 81 4.3.1. Double Integral of a of Two Variables ...... 81 4.3.2 Geometric Interpretation of the Double Integral...... 82 5. DIFFERENCE AND DIFFERENTIAL EQUATIONS ...... 84

5.1. BASIC IDEA OF A DIFFERENCE ...... 84 5.2. LINEAR DIFFERENCE EQUATIONS ...... 84 5.3. NUMERICAL EXAMPLE ...... 85 5.4. SOLUTION OF A FIRST-ORDER LINEAR DIFFERENCE EQUATION...... 85 5.4.1. Equilibrium Solution...... 85 5.4.2. General Solution ...... 86 5.4.3. Stability Analysis...... 87 5.5. SOLUTION OF A SECOND-ORDER LINEAR DIFFERENCE EQUATION...... 90 5.6. ORDINARY DIFFERENTIAL EQUATIONS ...... 91 5.6.1. A Simple Growth Model...... 91 5.6.2. First-Order Differential Equations...... 93 5.6.3. Second-Order Differential Equations...... 94 5.7. INTRODUCTION TO PARTIAL DIFFERENTIAL EQUATIONS...... 96 REFERENCES...... 99

3

CHAPTER ONE

Linear Algebra

4 1 Linear Algebra

1.1. What is a Matrix?

A matrix is a common device for displaying numbers or data in an organised form. Therefore a matrix can be viewed as a simple table of data. Using matrices to represent data makes the storage, displaying and manipulation of data quite convenient. For example, the test scores for a class of five students on three examinations. The best way to report these scores is to use the following table;

Student test 1 test 2 test 3 1 75 82 86 2 91 95 100 3 65 70 68 4 59 80 99 5 75 76 74

and in matrix form; ⎡75 82 86 ⎤ ⎢ ⎥ ⎢91 95 100⎥ test scores S = ⎢65 70 68 ⎥ ⎢ ⎥ ⎢59 80 99 ⎥ ⎣⎢75 76 74 ⎦⎥

• The number of columns and rows of the table, (5 x 3) in this case, are now the number of column and rows of the matrix, (5 x 3) and is known as the dimension of the matrix.

• Matrices are named by capital bold letters, A, S, etc.

• Each number in a matrix is called an element of that matrix and is denoted by the name of that matrix with a subscript noting the row and column of the element.

Therefore the general form of a matrix is

⎡a1,1 a1,2 L a1,n ⎤ ⎢ ⎥ a2,1 a2,2 L a2,n A = ⎢ ⎥ ⎢ ⎥ ⎢M O M ⎥ ⎣⎢am,1 am,2 L am,n ⎦⎥

A is a (m x n) matrix and ai,j represents the element in the ith row and jth column.

Examples: Identify the elements S1,3 , S2,2 , S5,3 , S4,1 , in the matrix S. S1,3 = 86 , S2,2 = 95 , S5,3 =74 , S4,1 =59

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1.2. Special types of matrices

There are special type of matrices according to the dimension, elements and the pattern of the elements inside the matrix.

1.2.1. Vectors

Vectors are special matrices with only one row or column. Therefore there are two types of vectors, a row vector and a column vector.

• A row vector, R, is a matrix which has dimension (1 x n).

R = [ r1,1 r1,2 . . . r1,n ]

As an example consider the test scores of the first student;

S1 = [ 75 82 86 ]

• A column vector, C, is a matrix which has dimension ( m x 1).

⎡c1,1 ⎤ ⎢ ⎥ c C = ⎢ 2,1 ⎥ ⎢ ⎥ ⎢M ⎥ ⎣⎢cm,1 ⎦⎥

As an example consider the test scores of the students in the first subject;

⎡75⎤ ⎢ ⎥ ⎢91⎥ S2 = ⎢65⎥ ⎢ ⎥ ⎢59⎥ ⎣⎢75⎦⎥

1.2.2. Square matrix

A matrix with equal number of rows and columns is known as a square matrix. ⎡ 1 3 5 ⎤ ⎡ 1 3 ⎤ ⎢ ⎥ A(1 x1) = [21] B(2 x2) = ⎢ ⎥ C(3 x3) = 7 9 11 ⎣ 5 7 ⎦ ⎢ ⎥ ⎣⎢13 15 17 ⎦⎥

6 1.2.3. Diagonal matrix

In a square matrix, the elements which have the same row and column index numbers (i.e. ai,j where i=j) are known as diagonal elements. Consequently, a matrix with non-zero diagonal elements and zeros elsewhere is called a diagonal matrix.

1.2.4. Identity Matrix

The Identity Matrix is a diagonal matrix in which the diagonal elements are equal to one and all off diagonal elements are zero.

⎡1 0 L 0⎤ ⎢ ⎥ 1 if ij= ⎢0 1 L 0⎥ ⎧ I = or aij, = ⎨ ⎢M O M ⎥ ⎩0 if ij≠ ⎢ ⎥ ⎣0 0 L 1 ⎦

If the upper diagonal elements of a square matrix equal to the lower diagonal elements, ai,j = aj,i , then the matrix is called symmetric.

⎡ 1 3 5 ⎤ Y = ⎢ 3 9 11 ⎥ ⎢ ⎥ ⎣⎢ 5 11 17 ⎦⎥

1.3. Matrix Operations

1.3.1 Adding and subtracting matrices

1) Two matrices can be added or subtracted if and only if they have same dimensions.

2) Only the corresponding elements are added together or subtracted from each other.

Therefore, for A (n × m) and B (n × m) ;

C(n × m) = A(n × m) + B(n × m) where c i,j = ai,j + b i,j C(n × m) = A(n × m) - B(n × m) where c i,j = ai,j - b i,j

⎡1 3⎤ ⎡2 4⎤ AB(2×× 2)= ⎢ ⎥ (2 2) = ⎢ ⎥ ⎣5 7⎦ ⎣6 8⎦

⎡1 3 ⎤ ⎡2 4 ⎤ ⎡1+ 2 3 + 4 ⎤ ⎡3 7⎤ C(2×2) = A(2×2) + B(2×2) = ⎢ ⎥ + ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎣5 7⎦ ⎣6 8⎦ ⎣5 + 6 7 +8⎦ ⎣11 15⎦ ⎡1 3 ⎤ ⎡2 4 ⎤ ⎡1- 2 3 - 4 ⎤ ⎡-1 -1⎤ C(2×2) = A (2×2) − B(2×2) = ⎢ ⎥ − ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎣5 7⎦ ⎣6 8⎦ ⎣5 - 6 7 - 8⎦ ⎣-1 -1⎦

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• Properties of matrix addition;

1) Commutative law A+B = B+A

2) Associative law (A+B)+C =A+(B+C)

These properties hold for matrix subtractions too, knowing that A-B=A+(-B).

1.3.2 Scalar multiplication

The product of matrix times a scalar is a new matrix in which each element is multiplied by the scalar

⎡ka1,1 ka1,2 L ka1,n ⎤ ⎢ ⎥ ka ka ka kA=Ak= ⎢ 2,1 2,2 L 2,n ⎥ ⎢ ⎥ ⎢ M O M ⎥ ⎣⎢kam,1 kam,2 L kam,n ⎦⎥

This process is known as scalar multiplication, since the matrix is scaled by the size of the scalar (number).

For example the test scores of the students can be scaled by 0.2, i.e. multiplying the score matrix by 0.2;

⎡75 82 86 ⎤ ⎡75(0.2) 82(0.2) 86 (0.2) ⎤ ⎢ ⎥ ⎢ ⎥ ⎢91 95 100 ⎥ ⎢91( 0.2) 95(0.2) 100(0.2)⎥ 0.2S = 0.2⎢65 70 68 ⎥ = ⎢65(0.2) 70(0.2) 68(0.2) ⎥ ⎢ ⎥ ⎢ ⎥ ⎢59 80 99⎥ ⎢59(0.2) 80(0.2) 99(0.2)⎥ ⎣⎢75 76 74 ⎦⎥ ⎣⎢75(0.2) 76 (0.2) 74(0.2) ⎦⎥

⎡15 16.4 17.2 ⎤ ⎢ ⎥ ⎢18.2 19 20 ⎥ = ⎢13 14 13.6 ⎥ ⎢ ⎥ ⎢11.8 16 19.8 ⎥ ⎣⎢15 15.2 14.8 ⎦⎥

1.3.3. Transpose of a Matrix

T The transpose of a matrix A, denoted by A’ (or A ), where A is (m × n) with elements ai,j , is T an (n × m) matrix with elements equal to aaij,,= ji.

8 ⎡a1,1 a 1,2 ⎤ ⎢ ⎥ ⎡a1,1 a 2,1 a 3,1 ⎤ AA'= a2,1 a 2,2 = ⎢ ⎥ ⎢ ⎥ a a a ⎢ ⎥ ⎣ 1,2 2,2 3,2 ⎦ ⎣a3,1 a 3,2 ⎦

⎡1 4 ⎤ ⎢ ⎥ ⎡1 3 5 ⎤ A = ⎢3 6⎥ A'= ⎢ ⎥ ⎣4 6 8⎦ ⎣⎢5 8⎦⎥

• The columns of AT are the rows of A and vice versa.

• The transpose of a row vector is a column vector and the transpose of a column vector is a row vector. • The transpose of a symmetric matrix is equal to the matrix itself. For example

⎡ 1 3 5 ⎤ ⎡ 1 3 5 ⎤ ⎢ ⎥ ⎢ ⎥ Y = ⎢ 3 9 11 ⎥ Y’ = ⎢ 3 9 11 ⎥ ⎣⎢ 5 11 17 ⎦⎥ ⎣⎢ 5 11 17 ⎦⎥

1) The transpose of a transposed matrix is equal to the original matrix: (A’)’=A 2) The transpose of a sum: (A+B)’= A’+B’ 3) The transpose of a product: (A B)’= B’A’

1.3.4. Vector multiplication

In order to compute the inner product of two vectors the number of elements in each vector have to be equal. The inner product of two vectors is a scalar.

This product is found by multiplying the individual elements of the row vector by the corresponding column elements of the column vector and adding them together.

Thus following the vector multiplication rule, the product of vectors R and C can be written as:

⎡c1,1 ⎤ ⎢ ⎥ c R = [ r r . . . r ] C = ⎢ 2,1 ⎥ 1,1 1,2 1,n ⎢ ⎥ ⎢M ⎥ ⎣⎢cn,1 ⎦⎥

9 ⎡c1,1 ⎤ ⎢ ⎥ c RC = r r r ⎢ 2,1 ⎥ = r c + r c + + r c []1,1 1,2 M 1,n ⎢ ⎥ []1,1 1,1 1,2 2,1 M 1,n n,1 ⎢M ⎥ ⎣⎢cn,1 ⎦⎥

Example; ⎡1⎤ ⎢ ⎥ ⎢2⎥ AB = [][2 4 6 8 10 ⎢3⎥ = (2×1) + (4× 2) + (6×3) + (8× 4) + (10×5) ] ⎢ ⎥ ⎢4⎥ ⎣⎢5⎦⎥ = [](2) + (8) + (18) + (32) + (50) =110

1.3.5 Matrix multiplication

In order to multiply two matrices, A and B to get AB, the necessary and sufficient condition is that the number of columns of the first matrix should be equal to the number of rows of the second matrix.

A × B (m × n) (p × q)

The matrices can be multiplied if and only if n=p

The product of two matrices can be computed using the inner product. This is done by computing the inner product of the ith row of A and jth column of B. This inner product will th represent the i,j element, ci,j , of the product matrix, C which will have dimension (m x q).

A × B = C

(m × n) (n × q) (m× q)

a) As an example, consider the following matrices

⎡1 3 2⎤ A = ⎡2 3 4⎤ B = ⎢ ⎥ ⎢ ⎥ 2 4 1 ⎣5 6 7 ⎦ ⎢ ⎥ ⎣⎢5 6 3⎦⎥

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Note that the number of columns of A is equal to the number of rows of B, therefore, we can perform the multiplication.

⎡21( )+ 32 ( )+ 45 ( ) 2(3) + 3(4) + 4(6) 2(2) + 3(1) + 4(3)⎤ CAB==⎢ ⎥ ⎣5(1) + 6(2) + 7(5) 5(3) + 6(4) + 7(6) 5(2) + 6(1) + 7(3) ⎦

⎡28 42 19⎤ = ⎢ ⎥ ⎣52 81 37⎦

It can be seen that the dimension of the product matrix is (2 x 3). b) For the second example consider the following matrices

⎡2 -3 ⎤ A = B = ⎡− 1 ⎤ (2×2) ⎢ ⎥ (2×1) ⎢ ⎥ ⎣-5 6 ⎦ ⎣ 2 ⎦

2 - 3 -1 2(−1) + (−3)(2) - 8 ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ C(2×1) = A (2×2)B (2×1) = ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎣- 5 6⎦ ⎣2 ⎦ ⎣(-5)(-1) + 6(2) ⎦ ⎣17 ⎦

• Product of a matrix with the identity matrix is the matrix itself

i.e. P(n×m) I(m×m) =P(n×m).

⎡ 1 3 2⎤ 1 0 0 ⎢-2 4 1⎥ ⎡ ⎤ P(4×3) = ⎢ ⎥ I(3×3)= ⎢ 0 1 0 ⎥ ⎢ 5 0 3⎥ ⎢ ⎥ ⎢ ⎥ ⎣⎢ 0 0 1 ⎦⎥ ⎣ 2 1 -1⎦

11 ⎡ 1 3 2⎤ ⎡ 1 0 0⎤ ⎢-2 4 1⎥ PI = ⎢ ⎥⎢ 0 1 0 ⎥ ()()43×× 33 ⎢ 5 0 3⎥⎢ ⎥ ⎢ ⎥⎣⎢ 0 0 1⎦⎥ ⎣ 2 1 -1⎦

⎡ 1(13020)()()++ 1(0) + 3(1) + 2(0) 1(0) + 3(0) + 2(1) ⎤ ⎢-2(1) + 4(0) +1(0) -2(0) + 4(1) +1(0) -2(0) + 4(0) +1(1) ⎥ = ⎢ ⎥ ⎢ 5(1) + 0(0) + 3(0) 5(0) + 0(1) + 3(0) 5(0) + 0(0) + 3(1) ⎥ ⎢ ⎥ ⎣2(1) +1(0) + (-1)(0) 2(0) +1(1) + (-1)(0) 2(0) +1(0) + (-1)(1)⎦

⎡ 1 3 2⎤ ⎢-2 4 1⎥ = ⎢ ⎥ = P ⎢ 5 0 3⎥ ()43× ⎢ ⎥ ⎣ 2 1 -1⎦

Note that it is also possible to multiply two vectors in the following way

⎡1⎤ ⎡2 4 6 8 10⎤ ⎢ ⎥ ⎢ ⎥ ⎢2⎥ ⎢4 8 12 16 20⎥ BA = ⎢3⎥[]2 4 6 8 10 = ⎢6 12 18 24 30⎥ ⎢ ⎥ ⎢ ⎥ ⎢4⎥ ⎢8 16 24 32 40 ⎥ ⎣⎢5⎦⎥ ⎣⎢10 20 30 40 50 ⎦⎥

1.3.6 Properties of matrix multiplication

1) The commutative law does not hold for matrix multiplication, however, the commutative law holds for scalar and matrix multiplication

AB≠BA kA=Ak

1) Associative law holds in matrix multiplication

(AB)C=A(BC)

2) Distributive law holds in matrix multiplication

A(B+C) =AB+AC

4) Commutative law holds when one of the matrix is the Identity matrix

AI=IA=A

12 1.4. The Determinant

For every square matrix, A(n×n) , a real number can be computed from the elements which is called determinant of that matrix and is denoted as |A| or DA . When the determinant of a square matrix is equal to zero, |A| = 0 , the matrix is called singular. When the determinant of a matrix is not zero, |A| ≠0, the matrix is non-singular.

The concept of determinant is quite important and useful in matrix operations and solving simultaneous equations.

The determinant of a (1 x 1) matrix is the value of the element inside the matrix. For example;

A=[20] , |A|=20 and M=[-2] , |M|=-2

1.4.1. How to find the determinant of a (2 x 2) matrix

For a (2 x 2) matrix of the following form

⎡a1,1 a 1,2 ⎤ A=⎢ ⎥ ⎣a21, a2,2 ⎦ the determinant of A can be computed by deducting the product of the elements on the secondary diagonal of A from the product of the elements on the principal diagonal of A.

Secondary diagonal

⎡a1,1 a 1,2 ⎤ A=⎢ ⎥ ⎣a21, a2,2 ⎦ Principal diagonal

Therefore, determinant of A can be computed by the following formula

a1,1 a 1,2

+ − A= =−aa aa 11,,, 22 12 21 a a 21, 2,2

As an example calculate the determinants of the following (2 x 2) matrices:

⎡2 - 2⎤ ⎡2 4 ⎤ a) A = ⎢ ⎥ ⇒ A = 2(1) - (-2)(3) = 8 ; b) B = ⎢ ⎥ ⇒ B = 2(12) - (4)(6) = 0 ⎣3 1 ⎦ ⎣6 12⎦

13 1.4.2. The determinant of a (3 x 3) matrix

The determinant of a (3 x3) matrix can be computed using the following procedure;

1) rewrite the first two columns of the matrix on the right hand side of the matrix. 2) add the products of the elements on the primary diagonals (P1 P2 P3 ). 3) add the products of the elements on the secondary diagonals (S1 S2 S3 ). 4) deduct the result from stage 3 from the result from 2.

S1 S2 S3

⎡a1,1 a 1,2 a 1,3 ⎤ a1,1 a 1,2 ⎢ ⎥ A= a a a a a ⎢ 2,1 2,2 2,3 ⎥ 2,1 2,2 ⎢ ⎥ ⎣a3,1 a 32 a 3,3 ⎦ a3,1 a 32 P1 P2 P3

Therefore, the determinant of a (3 x3) matrix can be formulated as follows;

A =+aa11,,, 22 a 33 aa 12 , 23 ,, a 31+ aaa 13 ,,, 21 32− aa 31 ,,, 22 a 13− a 32 , a 23 ,, a 11− aaa 33 ,,, 21 12

As an example, compute the determinant of the following (3 x 3) matrix;

⎡ 3 1 2 ⎤ S1 S2 S3 ⎢ ⎥ ⎡ 3 1 2 ⎤ 3 1 A = ⎢-1 2 4⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 3 - 2 1 ⎦ ⎢-1 2 4⎥ -1 2 ⎣⎢ 3 - 2 1 ⎦⎥ 3 - 2 P P P 1 2 3 The determinant of A is;

|A|=[(3)(2)(1)+ (1)(4)(3)+ (2)(-1)(-2)]-[(3)(2)(2)+(-2)(4)(3)+(1)(-1)(1)] = [(6+12+4)-(12-24-1)]=22-(-13)=35

1.4.3. Method of cofactors in determining determinants

A more general method of finding the determinant of a square matrix is the method of cofactors. For any square matrix A , there exists a matrix of cofactors with the same dimension as A which is denoted as Ac. For each element in A, ai,j , there is a corresponding element in Ac , a’i,j, which is called cofactor. In order to find the cofactor of the element (i,j) of matrix A, we follow this procedure:

1) Cross the ith row and jth column of the matrix. 2) Find the determinant of the remaining sub-matrix of the original matrix. 2) Call this determinant the minor of element (i,j).

14 4) Multiply the minor of the i,jth element by (-1)(i+j). This means that if i+j is even, then the sign assigned to the minor is positive. Similarly, if the sum of i+j is an odd number, then the sign assigned to the minor should be negative.

(i+j) The cofactor of ai,j is a’i,j =(-1) (minor i,j)

⎡2 -2⎤ As an example consider the (2 x 2) matrix A = ⎢ ⎥ ⎣3 1 ⎦

To find sub-matrix minor using formula cofactor ()11+ a’1,1 ⎡2 -2⎤ a'()()=−11 11, ⎢3 1 ⎥ ⎣ ⎦ 1 = (-1)2 ()11= 1 ()12+ 1,2 ⎡2 -2⎤ a'()()=−13 a’ 12, ⎢3 1 ⎥ ⎣ ⎦ 3 = (-1) 3 ()33=− -3 (2+1) 2,1 2 -2 a' = (−1) (−2) a’ ⎡ ⎤ 2,1 ⎢ ⎥ ⎣3 1 ⎦ -2 = (-1)3 (−2) = 2 2 (2+2) 2,2 ⎡2 -2⎤ a’ a'2,2 = (−1) (2) ⎢ ⎥ ⎣3 1 ⎦ 2 = (-1) 4 (2) = 2 2

⎡1 - 3 ⎤ Therefore, the matrix of cofactors for A can be written as; A c = ⎢ ⎥ ⎣2 2⎦

Let us find the matrix of cofactors for a (3 x 3) matrix, A,

⎡ 3 1 2⎤ ⎢ ⎥ A = ⎢-1 2 4⎥ ⎣⎢ 3 - 2 1 ⎦⎥

15 To find sub-matrix using formula cofactor ()11+ a’1,1 ⎡ 3 1 2⎤ a'()()11, =−110 ⎢ ⎥ 10 -1 2 4 2 ⎢ ⎥ = (-1) ()10= 10 ⎣⎢ 3 - 2 1 ⎦⎥ ()12+ a’1,2 3 1 2 ⎡ ⎤ a'()()12, =−113 − ⎢ -1 2 4⎥ 13 ⎢ ⎥ = (-1) 3 ()−=13 13 ⎢ 3 - 2 1 ⎥ ⎣ ⎦ ()13+ a’1,3 ⎡ 3 1 2 ⎤ a'()()13, =−14 − ⎢ -1 2 4⎥ -4 ⎢ ⎥ = (-1) 4 ()−=−44 ⎢ ⎥ ⎣ 3 - 2 1 ⎦ ()21+ a’2,1 ⎡ 3 1 2⎤ a'()()21, =−15 ⎢ ⎥ -5 3 ⎢ -1 2 4⎥ = (-1) ()55=− ⎣⎢ 3 - 2 1 ⎦⎥ ()22+ a’2,2 ⎡ 3 1 2⎤ a'()()22, =−13 − ⎢ ⎥ -3 4 ⎢ -1 2 4⎥ = (-1) ()−=−33 ⎣⎢ 3 - 2 1 ⎦⎥ ()23+ a’2,3 ⎡ 3 1 2⎤ a'()()23, =−19 − ⎢ ⎥ 9 -1 2 4 5 ⎢ ⎥ = (-1) ()−=99 ⎣⎢ 3 - 2 1 ⎦⎥ ()31+ a’3,1 ⎡ 3 1 2⎤ a'()()31, =−10 ⎢ ⎥ 0 -1 2 4 4 ⎢ ⎥ = (-1) ()00= ⎣⎢ 3 - 2 1 ⎦⎥ ()32+ a’3,2 ⎡ 3 1 2⎤ a'()()32, =−114 ⎢ ⎥ -14 -1 2 4 5 ⎢ ⎥ = (-1) ()14=− 14 ⎣⎢ 3 - 2 1 ⎦⎥ a’ ()33+ 3,3 ⎡ 3 1 2⎤ a'()()33, =−17 ⎢ ⎥ 7 -1 2 4 6 ⎢ ⎥ = (-1) ()77= ⎣⎢ 3 - 2 1 ⎦⎥

Therefore the matrix of cofactors for A(3 x3) is;

⎡ 10 13 -4 ⎤ A = ⎢ -5 -3 9⎥ c ⎢ ⎥ ⎣⎢ 0 -14 7 ⎦⎥

16 Example: How to find the determinant of the matrix using cofactors

1) Select one row or column of the matrix 2) Add up the product of the elements of the row (or column) by their corresponding cofactors. Therefore, selecting the first row of A, in the above example, the determinant of A can be computed as follows: |A| = [(3)(10) + (1)(13) + (2)(-4)]=[30+13-8] = 35 or using the first column of A: |A| = [(3)(10) + (-1)(-5) + (3)(0)]=[30+5+0] = 35

This general method can be used to find the determinant of any matrix with dimension greater than 2.

1.4.4. Properties of Determinants

• The determinant of a matrix will be zero, if all the elements of one row or column of that matrix are zero. ⎡0 0 0⎤ ⎡ 8 0 12⎤ ⎢ ⎥ ⎢ ⎥ A = ⎢2 4 1⎥ ⇒ A = 0 B = ⎢ 8 0 4 ⎥ ⇒ B = 0 ⎣⎢3 2 7⎦⎥ ⎣⎢12 0 28 ⎦⎥ • If two rows or columns of a matrix are interchanged, the sign of the determinant of that matrix will change. ⎡2 1 3 ⎤ ⎡2 1 3 ⎤ ⎢ ⎥ ⎢ ⎥ A = ⎢2 4 1⎥ ⇒ A = 17 B = ⎢3 2 7⎥ ⇒ B = −17 ⎣⎢3 2 7⎦⎥ ⎣⎢2 4 1⎦⎥ • If a row or a column of a matrix is multiplied by a scalar, the determinant of the matrix will be multiplied by that scalar too. ⎡2 1 3 ⎤ ⎡4 2 6⎤ ⎢ ⎥ ⎢ ⎥ A = ⎢2 4 1⎥ ⇒ A = 17 B = ⎢2 4 1⎥ ⇒ B = 2*17 = 34 ⎣⎢3 2 7⎦⎥ ⎣⎢3 2 7⎦⎥ • If any one column or row of a matrix is a linear combination of one or more columns or rows, then the determinant of the matrix is zero. ⎡2 1 3 ⎤ ⎢ ⎥ A = ⎢4 2 6⎥ ⇒ A = 0 ⎣⎢3 2 7 ⎦⎥ • If a multiple of a row or column of a matrix is added to another row or column of that matrix, the value of the determinant will not change.

• If a square matrix, A(n×n), is multiplied by a scalar, λ, then the determinant of the n matrix, B(n×n)=λA, will be; B = (λ )( A ) e.g. ⎡2 1 3 ⎤ ⎡ 8 4 12⎤ ⎢ ⎥ ⎢ ⎥ 3 A = ⎢2 4 1⎥ ⇒ A =17 B = 4*A = ⎢ 8 16 4⎥ ⇒ B = (4 )(17) =1088 ⎣⎢3 2 7⎦⎥ ⎣⎢12 8 28⎦⎥

17

1.5. Inverse of a matrix

1.5.1. Definition of an inverse matrix If the determinant of a matrix is not zero, i.e. the matrix is non-singular, there can be identified a matrix known as the inverse and is denoted as A-1 , with the following property.

AA−−11= A A= I

The inverse matrix is a unique square matrix with the same dimension as the original matrix.

As an example consider the following (2 x 2) matrices. ⎡2 3⎤ ⎡ 8 - 3 ⎤ A = ⎢ ⎥ B = ⎢ ⎥ ⎣5 8⎦ ⎣-5 2⎦

Let us find the product of A and B,

⎡2 3⎤⎡ 8 -3 ⎤ ⎡(2)(8) + (3)(−5) (2)(-3) + (3)(2) ⎤ AB = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎣5 8⎦⎣-5 2⎦ ⎣(5)(8) + (8)(-5) (5)(-3) + (8)(2)⎦

⎡(16) + (−15) (-6) + (6) ⎤ ⎡1 0⎤ = ⎢ ⎥ = ⎢ ⎥ = I ⎣(40) + (-40) (-15) + (16)⎦ ⎣0 1⎦

Therefore B is the inverse matrix of A.

1.5.2. How to find the inverse of a square matrix

The following procedure uses the matrix of cofactors to determine the inverse matrix;

1) In order to find the inverse of A, first find the matrix of cofactors, AC .

2) Using cofactors, find the determinant of the matrix, |A|.

3) If the determinant of the matrix is non zero, |A| ≠ 0, then find the adjoint matrix,

AA'jC= , which is the transpose of matrix of cofactors.

4) Dividing each element in the Adjoint matrix, Aj , by the determinant of the matrix, |A|, yields the inverse matrix , A-1.

Example: ⎡3 7⎤ Find the inverse of A = ⎢ ⎥ ⎣2 5⎦

18 1) matrix of cofactors ⎡ 5 - 2⎤ AC = ⎢ ⎥ ⎣-7 3 ⎦

2) determinant of A |A|=(3)(5)-(7)(2)=15-14=1

5 - 7 3) the Adjoint matrix ⎡ ⎤ A j ==A' C ⎢ ⎥ ⎣-2 3 ⎦ ⎡5 - 7⎤ ⎢ ⎥ -1 1 1 1 ⎡ 5 - 7 ⎤ 4) the inverse of A A = A = ⎢ ⎥ = | A | j - 2 3 ⎢- 2 3⎥ ⎢ ⎥ ⎣ ⎦ ⎣⎢ 1 1⎦⎥

5) verify the answer

−1 ⎡3 7⎤⎡ 5 - 7 ⎤ ⎡(3)(5) + (7)(−2) (3)(-7) + (7)(3) ⎤ ⎡1 0⎤ AA = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ = I ⎣2 5⎦⎣- 2 3⎦ ⎣(2)(5) + (5)(-2) (2)(-7) + (5)(3)⎦ ⎣0 1⎦

Example: ⎡ 0 1 2⎤ Find the inverse of B = ⎢-1 1 0⎥ ⎢ ⎥ ⎣⎢ 2 -1 3⎦⎥

1) the matrix of cofactors: 2) the adjoint matrix:

⎡ 3 3 -1⎤ ⎡ 3 - 5 - 2⎤ B = ⎢-5 - 4 2⎥ B = ⎢ 3 - 4 - 2⎥ C ⎢ ⎥ j ⎢ ⎥ ⎣⎢-2 - 2 1⎦⎥ ⎣⎢-1 2 1⎦⎥

3) the inverse of B:

⎡ 3 - 5 - 2 ⎤ ⎡ 3 - 5 - 2 ⎤ 1 1 B-1 = B = ⎢ 3 - 4 - 2 ⎥ = ⎢ 3 - 4 - 2 ⎥ | B | j 1 ⎢ ⎥ ⎢ ⎥ ⎣⎢-1 2 1⎦⎥ ⎣⎢-1 2 1⎦⎥

4) verify the answer:

⎡ 03 1 2⎤⎡ - 5 - 2⎤ ⎡ 1 0 0⎤ BB-1 = ⎢-1 1 0⎥⎢ 3 - 4 - 2⎥ = ⎢ 0 1 0⎥ =I ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣⎢ 2 -1 3⎦⎥⎣⎢-1 2 1 ⎦⎥ ⎣⎢ 0 0 1⎦⎥

1.6. Trace of a Matrix

Definition: Trace of a Matrix

19 The trace of a square NxN matrix is the sum of its diagonal elements

N

tr(A) = ∑aii Properties: i=1 1- tr(cA)=c(tr(A)), 2- tr(A')=tr(A) 3- tr(A+B)=tr(A)+tr(B)

4- tr(IN)=N 5- tr(AB)=tr(BA)

N N 6- tr(A'A)= a 2 ∑∑i==11j ij 7- tr(ABCD)=tr(BCDA)=tr(CDAB)=tr(DABC)

1.7. System of Linear Equations

A system of simultaneous equation equations can be written in matrix form. Ax=b

Where A is a known matrix of coefficients, x is a vector of unknowns and b is vector of constants.

Definition: Homogeneous Equation System A homogenous system is of the form Ax=0. A nonzero solution to such a system will exist if and only if A is singular; that is |A|=0.

Definition: Nonhomogeneous Equation System A nonhomogeneous system of equations is of the form Ax=b, where b is a nonzero vector. The solution to a nonhomogeneous system of equations exists and is unique if A is non- singular, i.e., |A|≠0. This means that the inverse of A exists and can be obtained. If we multiply both sides of the matrix equation by the inverse of the parameter matrix A-1 we can write: A-1Ax = A-1b Now we know that: AA-1 = I and Ix = x therefore we can write: A-1Ax = Ix = x = A-1b or x = A-1b Therefore, a system of simultaneous equations can be solved by pre-multiplying the inverse of the parameter matrix by the column vector of constants, if this matrix exists. The only condition for the system to have a solution is for the parameter matrix to be non- singular. In other words, the determinant of the parameter matrix should be different from zero.

Example Find the values of x and y, which satisfy the following system of equations

20 ⎧32xy+= 80 ⎨ ⎩xy+=240 the above equation can be written in a matrix form by separating the variables and parameters as follows;

⎡380 2⎤⎡x⎤ ⎡ ⎤ ⎢ ⎥⎢ ⎥ = ⎢ ⎥ or Ax=x ⎣1 2⎦⎣y⎦ ⎣40⎦ where A is a (2 x 2) matrix of parameters, X is (2 x 1) matrix of variables and B is (2 x 1) matrix of constants. The inverse of A can be determined as; 2 -2 ⎡ ⎤ ⎡ 22 -1⎤ ⎡ - 2⎤ 1 ⎢ 4 4 ⎥ AAA'A= == -1 ==A ⎢ ⎥ CjC⎢-2 3⎥ ⎢-1 3⎥ |A| j −1 3 ⎣ ⎦ ⎣ ⎦ ⎢ ⎥ ⎣⎢ 4 4⎦⎥ and the values of x and y

⎡⎛ 2 ⎞ ⎛ − 2 ⎞ ⎤ ⎜ ⎟(80) + ⎜ ⎟(40) ⎡2 - 2 ⎤ ⎢ 4 4 ⎥ ⎡(40) + (−20)⎤ ⎢ ⎥ ⎢⎝ ⎠ ⎝ ⎠ ⎥ −1 4 4 ⎡80⎤ ⎢ ⎥ ⎡20⎤ x = A b = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎢−1 3⎥⎣40⎦ ⎢ ⎥ ⎣10⎦ ⎛ −1⎞ ⎛ 3 ⎞ ⎣⎢(−20) + (30)⎦⎥ ⎣⎢ 4 4⎦⎥ ⎢⎜ ⎟(80) + ⎜ ⎟(40) ⎥ ⎣⎢⎝ 4 ⎠ ⎝ 4 ⎠ ⎦⎥

Example

Solve the following system of equations

⎧y + 2z = 25 ⎪ ⎨- x + y = 4 ⎪ ⎩2x - y + 3z = 27 We write the system in a matrix form ⎡ 025 1 2⎤⎡x⎤ ⎡ ⎤ ⎡ 0 1 2⎤ ⎡25⎤ ⎢-1 1 0⎥⎢y⎥ = ⎢4 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ A = ⎢-1 1 0⎥ b = ⎢4 ⎥ ⎣⎢ 2 -1 3⎦⎥⎣⎢z ⎦⎥ ⎣⎢27⎦⎥ ⎣⎢ 2 -1 3⎦⎥ ⎣⎢27⎦⎥

We compute the inverse of A, A-1 by the method of cofactors Matrix of cofactors ⎡ 3 3 -1⎤ A = ⎢-5 - 4 2⎥ determinant of A= |A|=1 C ⎢ ⎥ ⎣⎢-2 - 2 1⎦⎥

Adjoint matrix Inverse matrix ⎡ 3 - 5 - 2⎤ ⎡ 3 - 5 - 2⎤ A = ⎢ 3 - 4 - 2⎥ A=-1 ⎢ 3 - 4 - 2⎥ j ⎢ ⎥ ⎢ ⎥ ⎣⎢-1 2 1 ⎦⎥ ⎣⎢-1 2 1 ⎦⎥

21 Multiplying the inverse by the matrix of yields the solutions

⎡ 3 - 5 - 2 ⎤⎡25⎤ ⎡(3)(25) + (−5)(4) + (−2)(27)⎤ ⎡75 − 20 − 54 ⎤ ⎡1 ⎤ -1 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ X = A b = ⎢ 3 - 4 - 2 ⎥⎢4 ⎥ = ⎢(3)(25) + (−4)(4) + (−2)(27)⎥ = ⎢75 −16 − 54 ⎥ = ⎢5 ⎥ ⎣⎢-1 2 1⎦⎥⎣⎢27⎦⎥ ⎣⎢(−1)(25) + (2)(4) + (1)(27) ⎦⎥ ⎣⎢− 25 + 8 + 27⎦⎥ ⎣⎢10⎦⎥

1.8. Partitioned Matrices

Sometimes, it is useful to some of the elements in a matrix into sub-matrices as follows

⎛A11 A12 ⎞ A = ⎜ ⎟ ⎝A21 A22 ⎠

Where A is known as the partitioned matrix.

⎛1 4 5⎞ ⎜ ⎟ A = ⎜2 9 3 ⎟ ⎜ ⎟ ⎝8 9 6 ⎠ The determinant of a partitioned matrix A can be written as:

A A A = 11 12 = A * A − A A −1A = A * A − A A −1A 11 22 21 11 12 22 11 12 22 21 A 21 A 22

The inverse of the partitioned matrix can be obtained as follows:

−1 ⎛ A A ⎞ ⎛ A −1 (I + A F A A −1 ) − A −1A F ⎞ ⎜ 11 12 ⎟ = ⎜ 11 12 2 21 11 11 12 2 ⎟ ⎜ A A ⎟ ⎜ -1 ⎟ ⎝ 21 22 ⎠ ⎝ - F2 A 21A11 F2 ⎠

−1 −1 Where F2 = (A 22 − A 21A11 A12 ) .

Note that in order to apply the above operations for partitioned matrices, they must be conformable.

1.9. Vector spaces

R2 is a two dimensional space which is represented as the xy-plane using a coordinate 2 system. Each vector v=(v1,v2) in R has associated with it a unique point in the plane. The vector v can be represented geometrically by the arrow that begins at the origin and terminates at the point (v1,v2). The length or magnitude of v=(v1,v2) is defined to be the length of the arrow. 2 2 v = v1 + v1

22 10

8

6

4

2

0 -10-8-6-4-20246810 -2

-4

-6

-8

-10

• R2 is closed under scalar multiplication; every scalar multiple of a vector in the plane is also in the plane. • R2 is closed under addition; the sum of any two vectors in the plane is always a vector in the plane.

Definition: For each integer number n, we define the n-dimensional space Rn to be the set of all n-tuples (x1, x2,…,xn) of real numbers.

Definition: Magnitude of a vector in Rn n Let v=(v1,v2,…,vn) be a vector in R . The magnitude of v is 2 2 2 v = v1 + v2 + ... + vn

Definition: Orthogonal Vectors Two vectors v and w are orthogonal, v⊥w, if and only if their inner product is zero.

Definition: Parallel Vectors Two vectors v and w are parallel, v⎥⎥w, if one is a scalar multiple of the other, i.e, v=cw, with c in R.

Definition: Vector space A vector space is any set of vectors that is closed under scalar multiplication and addition, together with some properties that these operations must satisfied.

Theorem: Test for a subspace A nonempty subset W of a vector space V is a subspace if and only if it is close under vector addition and scaler multiplication.

Example R2 is a subspace of R3.

23 Theorem: Subspace generated or spanned by vectors n Let v1,v2,…,vn be vectors in a vector space R . The set

W = sp()v1,v2,…,vn = {r1v1 + r2v2 + ... + rnvn / ri ∈ R} of all linear combinations of these vectors is a subspace of V.

W is referred as the subspace of V generated or spanned by the vectors. The set {v1,v2,…,vn} is a spanning or generating set for V.

Example Any vector nonzero in R2 generates a one-dimensional subspace of R2

Definition: Linear Dependence A set of vectors is linearly dependent if any one of the vectors in the set can be written as a linear combination of the others.

Example Consider vectors a, b and c. These vectors are linearly dependent because any of them can be written as linear combination of the other two.

⎡1⎤ ⎡3⎤ ⎡10⎤ a = , b = , c = ⎢2⎥ ⎢3⎥ ⎢14⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 2a + b − 1 c = 0 or a = 1 c − 1 b b = 1 c − 2a 2 4 2 2

Definition: Linear Independence A set of vectors {v1,v2,…,vn}is linearly independent if

α v +α v +α v + ... +α v = 0 ⇒ α = 0∀i 1 1 2 2 3 3 k k i Example Any two non-parallel vectors in R2 are linearly independent.

Example Two orthogonal vectors are linearly independent.

Definition: Basis for a vector space A set of vectors in a vector space V is a basis for V if the following conditions are met • The set of vector generates V. • The set of vectors is linearly independent.

Example On a two-dimensional plane, if we have two vectors a and b non parallel, any other vector can be obtained as a linear combination of a and b. Consider vectors a and b in R2. We can find numbers α1 and α2 such that the arbitrary vector c can be written as c=α1 a+ α2b

⎡a ⎤ ⎡b ⎤ ⎡c ⎤ c1 = α1a1 +α 2b1 a = 1 , b = 1 , c = 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ c = α a +α b ⎣a2 ⎦ ⎣b2 ⎦ ⎣c2 ⎦ 2 1 2 2 2

24 a and b are a basis and the solutions to this pair of equation are

b c − b c a c − a c α = 2 1 1 2 , α = 1 2 2 1 1 2 a1b2 − b1a2 a1b2 − b1a2

this will have a unique solution provided a1b2 − b1a2 ≠ 0 . Therefore, any two vectors whose elements satisfy the condition a1b2 − b1a2 ≠ 0 can be used as basis and any other vector, c, can be formed as a unique linear combination of a and b.

Definition: Dimension The dimension of a vector space is the number of elements in any basis for V.

Observation If vectors a and b are a basis for R2 and c is another vector in R2, the space spanned by {a, b, c} is, again, R2.

Example Consider the set of three coordinate vectors whose third element is zero. For instance,

⎡a1 ⎤ ⎡b1 ⎤ a = ⎢a ⎥ , b = ⎢b ⎥ ⎢ 2 ⎥ ⎢ 2 ⎥ ⎣⎢0 ⎦⎥ ⎣⎢0 ⎦⎥

The vectors a and b do not span the three-dimensional space R3. Any linear combination of a and b has a third coordinate equal to zero, therefore they do not span the whole space R3. However the vectors, a and b, span a subspace of R3.

Theorem: Coordinate vector relative to a basis n Let B={b1,b2,…,bn} be a basis in the vector space R and let

v = r1b1 + r2b2 + ... + rnbn. The vector (r1,r2,…,rn) is the coordinate vector of v relative to the basis B.

1.9.1 Rank of a matrix

If we view a matrix as a set of column vectors, then the number of columns in the matrix equals the number of vectors in the set, and the number of rows equals the number of coordinates in each column vector.

Definition: Column Space The column space of a matrix is the vector space that is spanned by its column vectors.

For instance, ⎡1 5 6⎤ ⎢ ⎥ A = ⎢2 6 8⎥ ⎣⎢7 1 8 ⎦⎥

25 Matrix A contains three vectors from R3, however, the third is the sum of the first two. Thus, the column space of the matrix has dimension two.

Definition: The column (row) rank of a matrix is the dimension of the vector space that is spanned by its columns (rows).

Theorem: Equality of Row and Column Rank The column rank and row rank of a matrix are equal. We call it the rank of a matrix.

By the definition of row rank and its counterpart for column rank, we obtain the corollary

Corollary The row space and column space of a matrix have the same dimension.

Lemma • rank (A) = rank (A') ≤ min( number of columns, number of rows ) • rank(AB) = min(rank(A),rank(B)) • rank(A) = rank(A'A) = rank(AA')

Theorem: In a square matrix the rank is the number of linearly independent rows or columns in the matrix.

Definition: Matrix A(nxn) is full rank if rank(A)=n. Also matrix A(nxn) is reduced rank if rank(A)=r, where r

Theorem: The determinant of a matrix is nonzero (the matrix is non-singular) if and only if it has full rank.

1.10 Eigenvalues and Eigenvectors

Definition: Eigenvalues and eigenvectors Let A be an nxn matrix. A scalar λ is an eigenvalue of A if there is a non zero vector v in Rn such that Av=λv. The vector v is then an eigenvector of A corresponding to λ (the terms characteristic vector and characteristic value are also used).

We may chose v with length one, i.e. v′v=1, to remove indeterminacy. The solution is then λ and the n-1 unknown elements in v.

In principle, equation Av=λv or Av=λIv can be solved as follows

Av=λIv ⇒ (A- λI)v=0

This is a homogenous system that has a nonzero solution only if the matrix (A- λI) is singular i.e has a zero determinant. Therefore, if λ is a solution,

26

A − λI = 0

This in λ is known as the characteristic equation of A. For a (nxn) matrix, the characteristic equation is an nth-order polynomial in λ. The solution may be n real or complex values that may be distinct or repeated.

Example If ⎛5 1⎞ A = ⎜ ⎟ ⎝2 4⎠ Then

5 − λ 1 2 A − λI = ⇒ (5- λ)(4- λ) - 2*1= 0 ⇒ λ − 9λ +18 = 0 2 4 - λ

The characteristic roots of A are the solutions to the above equation; that is, λ1=6 and λ2=3.

2 − λ 0 = (2 − λ)2 ⇒ λ = λ = 2 0 2 - λ 1 2

1− λ 0 = λ2 − 5λ = 0 ⇒ λ = 5 and λ = 0 0 4 - λ 1 2

With λ in hand, the characteristic vectors are derived from the original problem, Ac=λv or (A-λI)v=0

5 − λ 1 ⎛v ⎞ 0 ⎛ ⎞ 1 ⎛ ⎞ ⎜ ⎟⎜ ⎟ = ⎜ ⎟ ⎝ 2 4 - λ ⎠⎝v2 ⎠ ⎝0⎠

Using the two value derived for λ (λ1=6, λ2=3) we can write two sets of simultaneous equations as follows

For λ=6 -v1+v2=0 2v1-2v2=0

for λ=3 2v1+v2=0 2v1+v2=0

Imposing the additional condition v’v=1 we find:

1 ⎛ 1 ⎞ ⎛ 2 ⎞ 5 For λ = 6 ⇒ v = ⎜ ⎟ and For λ = 3 ⇒ v = ⎜ ⎟ ⎜ 1 ⎟ ⎜− 2 ⎟ ⎝ 2 ⎠ ⎝ 5 ⎠

27

1.10.1 General results for characteristic roots and vectors

Theorem: A (nxn) symmetric matrix has n distinct characteristic vectors, c1, c2, ..., cn. The corresponding characteristic roots λ1, λ2, ..., λn are real but not necessarily distinct.

Theorem: The characteristic vectors of a symmetric matrix are orthogonal, i.e. ci′cj=0 for every i≠j.

Diagonilisation of a Matrix: th The characteristic vectors can be collected in (nxn) matrix C, in which the i column is ci corresponding to λi.

C = (c1 c2 ... c N )

Similarly the n characteristic roots (eigenvalues) can be collected in the following diagonal matrix.

⎛λ 0 0 ⎞ ⎜ 1 L ⎟ ⎜0 λ 0 ⎟ Λ = 2 ⎜ 0 ⎟ ⎜ M O ⎟ ⎜0 0 λ ⎟ ⎝ L N ⎠

Then we can write the full set of equations Aci=λici i=1,2, … n in the form AC=CΛ (*)

' On the other hand, since the eigenvectors are orthogonal (cic j = 0 for i ≠ j) and of length one (c’c=1), it can be seen that C'C=I and thus C'=C-1.

⎛c' c c' c c' c ⎞ ⎜ 1 1 1 2 L 1 N ⎟ ⎜ ' ' ' ⎟ c2c1 c2c2 L c2c N C'C = ⎜ ⎟ = I ⎜ M O M ⎟

⎜c' c c' c c' c ⎟ ⎝ N 1 N 2 L N N ⎠

Multiplying both sides of (*) by C’ we find that C'AC=C'CΛ=Λ

We say that matrix A can be diagonalised.

28 1.10.2 Further results on determining the rank of a matrix

Note: If A can be diagonalized the rank of A is equal to the rank of Λ. Also since Λ is a diagonal matrix, its rank is just the number of nonzero values on its diagonal.

Theorem: Rank of a symmetric matrix The rank of a symmetric matrix is equal to the number of nonzero characteristic roots of the matrix.

Theorem: Rank of a matrix The rank of a matrix is equal to the number of nonzero characteristic roots of A'A.

Theorem: Determinant of a matrix. If A can be diagonalized, its determinant is equal to the product of its characteristic roots.

We know that C’AC= Λ, therefore, |C’AC|=|Λ|.

|C’AC|= |C’| . |A| . |C|= |C’| . |C|. |A|= |C’C| . |A| = |I| . |A|=1 . |A| = |A| = |Λ|

Theorem: Characteristic roots of the Inverse Matrix. If A-1 exists, the characteristic roots of A-1 are the reciprocal of those of A, and the characteristic vectors are the same. ⎛λ 0 0 ⎞ ⎛1/ λ 0 0 ⎞ ⎜ 1 L ⎟ ⎜ 1 L ⎟ ⎜0 λ2 0 ⎟ -1 ⎜0 1/λ2 0 ⎟ for A ⇒ Λ = , A ⇒ Λ −1 = A ⎜ 0 ⎟ A ⎜ 0 ⎟ ⎜ M O ⎟ ⎜ M O ⎟ ⎜ ⎟ ⎜ ⎟ ⎝0 0 L λN ⎠ ⎝0 0 L 1/λN ⎠

Theorem: Trace of a Matrix of the Inverse Matrix. If A can be diagonalised the trace is the sum of its characteristic roots.

N tr(A)= λ ∑i=1 i

1.11 Definite Matrices

• Definition: Positive definite matrix. Matrix A is said to be positive definite if x'Ax>0 for all nonzero x.

• Definition: Negative definite matrix. Matrix A is said to be negative definite if x'Ax<0 for all nonzero x.

• Definition: Positive semi-definite (nonnegative) matrix. Matrix A is said to be positive semi-definite if x'Ax≥0 for all nonzero x.

29

• Definition: Negative semi-definite (nonpositive) matrix. Matrix A is said to be negative semi-definite if x'Ax≤0 for all nonzero x.

Theorem: Definite Matrices If all the characteristic roots of A are positive, then A is positive definite, while if all the characteristic roots of A are negative, then A is negative definite. Also, if some of the roots are zero and the remaining are positive (negative), then A is nonnegative (nonpositive) definite. If A has both negative and positive roots, then A is said to be indefinite.

The following are some useful results

• If A is nonnegative definite, then A ≥ 0 ,

• If A is positive definite, so is A-1,

• The identity matrix is positive definite,

• If A is positive definite and B is a non-singular matrix, then B'AB is positive definite.

30

CHAPTER TWO

Differential Calculus (The one case)

31 2. Differential Calculus

2.1. Differentiation

2.1.1. Definitions

Definition: The limit of f(x) as x approaches a equals L if we can make the values of f(x) arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a. We write limf (x) = L x→a

Definition: A function f is continuous at a number a if limf (x) = f (a) x→a Definition: of a line The slope of a line is define as the amount by which its vertical co-ordinate changes for a unit increase along the horizontal axis, i.e. change in the dependent variable slope = unit change in the indep. variable Geometrically, the slope of a line is just the of the angle created by the intersection of the line and the horizontal axis.

Definition: Average rate of change The average rate of change of y=f(x) with respect to x over the [x1,x2] is the difference quotient ∆y f (x )− f (x ) = 2 1 , ∆x x2 − x1 and can be interpreted as the slope of the through the points P(x1,f(x1)) and Q(x2,f(x2)) .

Definition: Instantaneous rate of change The instantaneous rate of change of y=f(x) with respect to x at x=x1, which is interpreted as the slope of the tangent to the curve y=f(x) at P(x1,f(x1)), is define as

∆y f (x )− f (x ) lim = lim 2 1 x →x x →x 2 1 ∆x 2 1 x2 − x1

Definition: The derivative The derivative of a function f at a number a, is f (x)− f (a) f (a + h)− f (a) f ' ()a = lim = lim x→a x − a h→0 h provided this limit exists.

Continuity is a necessary condition for differentiability but it is not sufficient. That is, although differentiability implies continuity, the converse it is not true.

32 Continuity at a point only rules up the presence of a gap, whereas differentiability rules out “sharpness” as well. Therefore, differentiability calls for “smoothness” of the function as well as its continuity.

Example: The function y = f (x) =| x − 2 | +1 is continuous at x=2,but it is not differentiable at x=2.

Observations: 1. The tangent line to the curve y=f(x) at P(a,f(a)) is the line through P whose slope is f’(a) , the derivative of f at a. 2. The derivative f’(a) is the instantaneous rate of change of y=f(x) with respect to x when x=a. 3. The derivative f’(x) is itself a function 4. If y=f(x) is a of x, the concepts of average rate of change and the instantaneous rate of change (the derivative) are equivalent. In other words, the tangent line to a straight line is the line itself. Also the slope of a line is constant.

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0 13579 -5

The tangent line as limit of secant lines It can be seen that as ∆x gets smaller and smaller the secant lines becomes the tangent line to the curve at point A.

Example: rate of change of linear function

The production function of a company has the following form:

yx=+32

Where, y is the output and x is the input, find change in the production for a unit change in input.

1) Choose two points on the domain, and find the respective function values

33 yfx==+() 32 x

x11 = 1 y =+=31() 2 5

x22 = 2 y =+=32() 2 8

2) Compute the slope:

∆y y − y 85− 3 m ==21= ==3 ∆x xx21− 21− 1

The rate of change in the company’s production function is 3. This means that if the input for this company increase by one unit the production will increase three times.

Example: rate of change of non linear function

The revenue of a company is given by the non-linear form: yxx=+510 −2

Where x represents the units of output (for example, cars).

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10 1357

The rate of change in the revenue (slope of the tangent to the curve) is not the same at different levels of output. It increases in the beginning and reaches a maximum, and declines after the maximum.

If we plot the graph of the different values taken by the tangent on the curve, we find:

34 15

10

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0 -11357911

-5

-10

Plot of the derivative function of the revenue

The plot shows that the rate of change of the function is positive for values of output lower than 5 and negative for values of output bigger than 5.

2.1.2. Rules of Differentiation

1) The Rule Derivative of a constant is zero, fx()= c

yfx''()= = 0

e.g. yfx==() 4 yfx''()= = 0

2) The Power Function Rule Derivative of an exponent, fx()= xn , is:

n−1 yfxnx''()==

e.g. a) yfx==() 2 x2 yfx''()= = 4 x b) yfx==() 3 x6 yfx''()==18 x5 1 c) yfx==() x3 yfxx''()==2 3

3) Constant times a Function Rule

fx()= cgx .() Derivative of a constant times a function is equal to the constant times the derivative of that function: yfxcgx'= '()= .'()

35 4) The Sums of Functions Rule Derivative of sums and differences of two functions fx()= gx ()± hx (), is:

yfxgxhx'= '()= '()± '()

e.g. a) yfx==−() 2 x25 3 x yfx''()==−415 x x4 1 52 b) yfx==+() 3 x63 x yfx''()==+18 x x 3

5) Derivative of product of two functions fx()= gxhx ().(), is:

yfxgxhxgxhx'= '()= '().()+ ().'()

e.g. a) yfx==()213 x2 ( − x ) yx'(=−+−=−−=−413 x )()() 32 x2222 4 x 12 x 6 x 4 x 18 x

1 b) yfx==+() (23 x23 ) x 3 1 yxx'()(=++=++=+6 3224244223 xxx ) 22352 x x xx 3

6) The derivative of the ratio of two functions, f(x)= g(x) / h(x) is:

g'(x)h(x) − h'(x)g(x) y'= f '(x) = [h(x)]2 Where g(x) is a

3x 2 − x + 3 e.g y = x 3 −1 (x 3 −1)(6x −1) − (3x 2 − x + 3)(3x 2 ) y'= f'(x) = (x 3 -1) 2 [6x 4 − x 3 − 6x +1] −[9x 4 − 3x 3 + 9x 2 ] y'= f'(x) = (x 3 −1) 2

7) The Power of a Function Rule The derivative of a function raised into the power of n, fx()= [()] gxn is:

n−1 yfxngxgx''()[()]'()==

Where g(x) is a differentiable function

36

e.g. a) yfx==−() (13 x23 ) gx()=−13 x2 g '(x ) = −6 x y' = 3(1 − 3x 2 ) 2 (−6 x) = −18 x(1 − 3x 2 ) 2 = (−18 x)(1 − 6 x 2 + 9 x 4 ) = -162x 5 + 108 x 3 − 18 x

8) If, yfu= ( ) and ugx = ( ) , where g(x) and f(x) are both differentiable functions, then the of derivatives can be applied as follows:

dy dy du y'= = ⋅ dx du dx

e.g. a) y = (x 2 + 3x −1)3 let us assume y = u 3 and u = x2 + 3x − 1 dy dy du d(u 3 ) d(x 2 + 3x −1) = ⋅ = ⋅ dx du dx du dx

= 3u 2 ⋅(2x + 3) = 3(x 2 + 3x −1) 2 (2x + 3)

b) y = (x3 − 3) 2 let us assume y = u 2 and u = x3 − 3 dy dy du d(u2 ) d(x3 − 3) = ⋅ = ⋅ dx du dx du dx

= 2u ⋅ (3x2 ) = 2(x3 − 3)(3x2 ) = 6x5 −18x2

2.1.3. Derivatives of Exponential and logarithmic functions

1) The Natural Rule

i) f (x) = e x f'(x) = e x ii) f(x) = e g(x) f'(x) = g'(x)e g(x)

iii) f(x) = ke g(x) f'(x) = kg'(x)e g(x) iv) f(x) = a g(x) f'(x) = g'(x)a g(x) .ln a

e.g. a) y = e 2 x y' = 2e 2 x 3 2 3 2 b) y = e x − x y' = ( 3x 2 − 2x)e x − x 2 2 c) y = e 3(x −1 ) y' = 6xe 3(x −1 )

37

2) The Natural Logarithmic Function Rule

1 i) f(x) = ln x f'(x) = x g ' ( x ) ii) f(x) = ln g ( x ) f'(x) = g ( x )

iii) f(x) = log a g ( x ) ln g ( x ) g ' ( x ) = f'(x) = ln a g ( x ) ln a

36 x 2 3 e.g. a) y = ln 12 x 3 ⇒ y' = = 12 x 3 x 2x − 1 b) y = ln( x 2 − x + 1) ⇒ y' = x 2 − x + 1 5 1 c) y = ln(5x) ⇒ y' = = 5x x 2x 1 d) y = log( x 2 + 4) ⇒ y' = * x 2 + 4 ln 10

2.1.4. Higher order derivatives

It is possible to find the derivative of a derivative of a function. This is called the of a function and measures the slope and the rate of change of the first derivative.

The notation for the second derivative is y’’ or f’’(x). Similarly, the of a function, f’’’(x), measures the slope and the rate of change of the second derivative and so on and so forth. The rules for taking the second and higher derivatives are the same as before.

Examples:

a) yfx==−() 2 x25 3 x yfx''()==−415 x x4 yfx''==− ''( ) 460 x3 y'' ' = f '' '( x) = −180 x 2

1 b) yfx==+() 3 x63 x yfx''()==+18 x52 x 3 yfx''==+ ''( ) 90 x4 2 x y' ' ' = f ' ' ' ( x) = 360 x 3 + 2

38

2.2. Applications of Derivatives

So far we learned how to find the derivative of a function. Now we study the use of these derivatives in applications such as plotting functions, finding maximum, minimum, point of inflection and of a function. We also study some economic applications of derivatives.

2.2.1. Taylor and Maclaurin Series

dy d{f (x)} Let y = f(x), then ≡ ≡ f 1(x) dx dx d 2 y d{f 1(x)} ≡ ≡ f 2(x) dx2 dx d 3 y d{f 2(x)} ≡ ≡ f 3(x), etc . dx3 dx

Taylor's Series

A function, f(x), may be approximated at some point, xo , by the following: 1 f(x) = f(x ) + f 1 (x ) (x - x ) + f 2 (x ) (x - x )2 0 0 0 2! 0 0

1 1 + f 3 (x ) (x - x )3 + + f n(x ) (x - x )n + Remainder 3! 0 0 K n! 0 0 where f(x0) is the value of f(x) at the point x0 and n! reads n where n! = n × (n - 1) × (n - 2 ) K 2 × 1 . Note 0! = 1! = 1.

Maclaurin's Series

A special case of the is whenx0 = 0 , in which case 1 1 1 f(x) = f( 0 ) + f 1 ( 0 )x + f 2 ( 0 ) x 2 + f 3 ( 0 ) x3 + + f n ( 0 )x n + Remainder 2! 3! K n!

Examples of Maclaurin series expansion

Example 1: Let f(X) = (1 + X)3 We know that (1 + X)3 = 1 + 3X + 3X 2 + X 3 . Use MacLaurin's Series to prove it.

f(X) = (1 + X)3 f( 0 ) = 1 f 1(X) = 3(1 + X)2 f 1( 0 ) = 3 f 2(X) = 6(1 + X) f 2( 0 ) = 6

39 f 3(X) = 6 f 3( 0 ) = 6 f 4 (X) = 0 f 4 ( 0 ) = 0

1 1 ∴ f(X) = 1 + 3 X + 6 X 2 + 6 X 3 2! 3! Thus (1+ X)3 = 1 + 3 X + 3 X 2 + X 3

Example 2: Find the value of e Let f(X) = e x then f(X) = e x f( 0 ) = e0 = 1 f 1(X) = e x f 1( 0 ) = e0 = 1 f 2(X) = e x f 2( 0 ) = e0 = 1 M M

X 2 X 3 X n ∴ e x = 1 + X + + + + + ... 2! 3! K n

Another example of Taylor Series

Taylor series expansion is used to measure the sensitivity of bond prices with respect to changes in interest rates. Using the first two terms of a Taylor series expansion to approximate the price change we can write

dP 1 d 2 P dP = dy + (dy)2 + error dy 2 dy 2

Dividing both sides of equation by P to find the percentage price change results in

2 dP dP 1 1 d P 1 2 error = dy + 2 (dy) + P dy P 2 dy P P The duration of the bond is define as 1 dP D = − . P dy The convexity of the bond is defined as d 2 P C = dy 2

40 Bond Price

Actual price increase The line representing slope of the curve

Actual price fall

Current rate Interest rate Rate decrease Rate increase

2.2.2. Increasing and decreasing functions

Definition: A function f(x) is said to be increasing (or monotonically increasing), if successively larger values of the independent variable x always lead to successively larger values of f(x), that is if x1>x2 ⇒ f(x1)>f(x2)

On the other hand, if successive increase in x always lead to successive decrease in f(x), that is, x1>x2 ⇒ f(x1)

then the function is said to be decreasing (or monotonically decreasing) .

Proposition: If a function is differentiable and decreasing at an interval, its first derivative is negative, f’(x)<0 at this interval. This indicates that the slope of the curve at any point, within that interval, is negative. The figure below illustrates a decreasing function, which has a negative first derivative at point A and within the chosen interval.

41 ∆x 0.8

0.7

0.6

∆y0.5

0.4

0.3

0.2

0.1

0 -1 -0.8 -0.6 -0.4 -0.2 0 0.2

[Decreasing function]

Proposition: By the same token, a function with a positive first derivative within an interval is an increasing function. The slope or value of the tangent to any point on an increasing function is positive. The following figure shows an increasing function

0.8

0.7

0.6

∆y 0.5

0.4

0.3

∆x 0.2

0.1

0 -1.8 -1.6 -1.4 -1.2 -1

[Increasing function]

2.2.3. Concavity and Convexity

Definition: A function is said to be concave at x=a, if at some small region close to the tangent to the line at point [x=a, f(a)], the graph of the function lies below the tangent line. On the other hand a function is said to be convex at x=a, if at some small region around the point [x=a,f(a)], the graph of the function lies completely above the tangent line. The figures below show the graphs of two concave and convex functions.

42 -0.1 0.8 1 1.2 1.4 1.6 1.8 2 -0.2 0.7 0.6 -0.3 0.5 -0.4 0.4 -0.5 0.3 0.2 -0.6 0.1 -0.7 0 -0.8 -1.8 -1.6 -1.4 -1.2 -1

a) A convex function b) A

The first derivative measures the rate of change of a function. The second derivative is the measure of the rate of change of the first derivative. We can distinguish the following cases (see the figures): 1. If ' '' f (x0 ) > 0 and f (x0 ) > 0 then the value of the function at x0 is increasing at an increasing rate. 2. If ' '' f (x0 ) > 0 and f (x0 ) < 0 then the value of the function at x0 is increasing at a decreasing rate. 3. If ' '' f (x0 )< 0 and f (x0 ) > 0 then the slope at x0 is negative and increasing. 4. If ' '' f (x0 )< 0 and f (x0 )< 0 then the slope at x0 is negative and decreasing.

The first derivative a function f tells us about its slope. The second derivative of a function informs us about the curvature of its graph. We can find whether a function is concave or convex at a point by finding the value of its second derivative at that point.

Proposition: If at x=a y''=f''(a)>0 then f(x) is convex at x=a

If at x=a y''=f''(a)<0 then f(x) is concave at x=a

43

a) f’(x)>0 b) f’(x)<0 f’’(x)>0 f’’(x)>0

c) f’(x)>0 d) f’(x)<0 f’’(x)<0 f’’(x)<0

Examples: Find the slope and curvature of following functions at x=3;

a) y = −x 2 + 1 yfx''()==−230 x f ' ( ) = -6 <

yfx''==− ''( ) 2 f' ' (3) = -2 < 0 Therefore, the function has a negative slope and is concave at x=3.

b) y = x 3 + x y'= f '(x) = 3x 2 +1 f '(3) = 28 > 0

y'' = f''(x) = 6x f''( 3)=18>0 Therefore, the function has a positive slope and is convex at x=3.

2.2.4. First derivative test for relative extremum

At a relative maximum or minimum point the function is neither decreasing nor increasing. Therefore if a relative extremum occurs, either the first derivative does not exist or if it does is zero. We restrict ourselves to continuous and differentiable function. Thus ' f (x0 ) = 0 it is a necessary condition for a relative extremum. However it is not a sufficient condition. The points of the domain at which the first derivative of the function is zero, are called critical values. For example, in the figure below, the slope of the tangent lines at x= -1 and

44 x=1 are zero and at these two point there are a relative maximum and a relative minimum, respectively.

2.2.5. Second derivative test for relative extremum

If x=x0 is a critical value of f(x), then f(x0) will be 1. A relative maximum if f’(x) changes its sign from positive to negative from the immediate left of the point x0 to its immediate right. Equivalently, If f’’(x0)<0, then the point [x=x0, f(x0 )] is a maximum 2. A relative minimum if f’ (x) changes its sign from negative to positive from the immediate left of the point x0 to its immediate right. Equivalently, If f’’(x0)>0, then the point [x=x0, f(x0 )] is a minimum

Note: In order to find maximum and minimum points on a curve (function), 0.8 we follow two steps; 0.6 i) Take the first derivative of the 0.4

function, f’(x), set it equal to 0.2 zero and solve for the critical point(s), x=x1 , x=x2 , etc. This 0 step is known as the first-order -2 -1 0 1 2 -0.2 condition, FOC. -0.4 ii) Take the second derivative of -0.6 the function, f’’(x), and evaluate it at the critical point(s). This -0.8

step is known as second-order [Relative maximum and minimum] condition, SOC.

Example: Find relative maximum and minimum points of the following functions; a) yx=−2 +1 FOC : y'= f '(x) = −2x − 2x = 0 ⇒ x = 0

SOC : y'' = f ''(x) = −2 f ''(x) = −2 < 0 Therefore, the function has a relative maximum at [x=0 , y=1]. a) y = 2x 3 − 6x + 12 2 2 FOC : y'= f '(x) = 6x - 6 6x − 6 = 0 ⇒ x1 = 1 , x2 = −1

SOC : y'' = f ''(x) = 12x f ''(1) = 12 > 0 f ''(−1) = −12 < 0 Therefore, the function has a relative maximum at [x=-1, y=16] and a relative minimum at [x=1, y=8].

45

2.2.6. The Nth-derivative test

If f’(x0)=0 and f’’(x0)=0 then f(x0) can be either a relative maximum or minimum, or an inflectional value (an inflectional value is a point of the graph in which the curvature changes from concave to convex or vice-versa). In this case we proceed with the n-th derivative test. If ' '' n−1 n f (x0 ) = 0, f (x0 ) = 0,...., f (x0 ) = 0, f (x0 ) ≠ 0 then the critical value f(x0) will be n • A relative maximum if n is an even number and f (x0)<0 n • A relative minimum if n is an even number and f (x0)>0 • An inflection point if n is an odd number.

2.2.7. Economic applications of derivatives

2.2.7.1. Relationship between total, marginal, and average concepts

• The sum of money that flows into a firm is referred to as revenue. • The sum of money, which flows out of a firm is known as costs. There are different ways to define revenue and costs.

2.2.7.2. Marginal and Average Revenue

• Total Revenue, TR, is defined as the revenue derived from all sales of particular goods and is the product of the total number of unit sold, Q, and the unit price of the goods, P. Mathematically,

TR = PQ⋅

• Marginal Revenue, MR, is defined as the change in the Total Revenue caused by a unit increase in production. Therefore, using derivative notation

MR = d(TR) d(Q)

• Average revenue, AR, is defined as the revenue from the sale of each unit, and can be obtained by dividing the total revenue by the number of units sold as follows;

TR PQ AR= = = P Q Q Find the average and marginal revenues for the following total revenue function and evaluate them at Q=10 and Q=5. TR = 200Q − 3Q 2

46 d(TR) TR 200Q − 3Q 2 MR = = 200 − 6Q AR = = = 200 − 3Q dQ Q Q

If Q = 10 MR = -6(10) + 200 = 140 If Q = 10 AR = -3(10) + 200 = 170 If Q = 5 MR = -6(5) + 200 = 170 If Q = 5 AR = -3(5) + 200 = 185

2.2.7.3. Marginal and Average Cost (cost minimisation)

• Total Cost, TC, is defined as the total cost of all factors of production involved in the production of a certain level of output. Total cost consist of two parts, Total Fixed Cost (TFC) and Total variable Cost (TVC). Where, Fixed cost are those which are constant and do not vary with the output level, while, the variable cost are those which vary with different levels of output. Therefore we can write;

TC=f(Q)=TVC+TFC

• Marginal Cost, MC, can be defined as the change in the total cost caused by producing an additional unit of output. d(TC) MC ==fQ'( ) d(Q)

• Average Cost, AC, is the cost of production of one unit which can be obtained by dividing the total cost by the number of units produced in the following form; TC FQ() AC == Q Q

Example 1: Find the average cost, AC, and marginal cost, MC, for the following total cost, TC, function and evaluate them at Q=10 and Q=5.

TC = Q 3 − 10 Q 2 + 30 Q + 20

d(TC) TC Q3 −10Q2 + 30Q + 20 MC = = 3Q2 − 20Q + 30 AC = = = Q2 −10Q + 30 + (20/Q) dQ Q Q

2 2 If Q = 10 MC = 3(10) - 20(10)+ 30 = 130 If Q = 10 AC = (10) -10(10)+ 30 + (20/10)= 32

If Q = 5 MC = 3(5)2 - 20(5)+ 30 = 5 If Q = 5 AC = (5)2 -10(5)+ 30 + (20/5) = 9

Now, if we sketch the graphs of total and marginal cost curves, we see that;

1) Total cost first rises until reaches a maximum and then falls until reaches its minimum and then rises again.

47 2) Marginal cost curve is positive until the total cost reaches its maximum, at this point marginal cost becomes negative indicating that total cost is falling. Marginal costs stays at the negative region until total cost reaches its minimum. At this point, marginal cost will become positive indicating that the total cost is rising again.

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35 Cost

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-5 0123456789101112 Quantity

TC MC

Now let us find the minimum and maximum of the total cost function. This is equivalent to setting the MC to zero and solve it to find where the MC cuts the horizontal axis.

2 20 ± 400 − 4*3*30 MC = 3Q − 20Q + 30 = 0 ⇒ Q1,Q2 = 6 20 + 40 20 − 40 ⇒ Q = = 4.38 and Q = = 2.79 1 6 2 6

2.2.7.4. Profit Maximization

Profit, Π, of an organisation is the total revenue less total cost and can be shown as;

⎪⎧TC = 4Q 2 −12Q +10 Π=TR − TC Where ⎨ 2 ⎩⎪TR = 3Q + 8Q + 5 Therefore, using the cost and revenue functions above, the profit function will be;

Π = TR −TC

Π = f (Q) = (3Q 2 + 8Q + 5) − (4Q 2 −12Q +10) = −Q 2 + 20Q − 5

Having the firm’s total cost and revenue functions (above examples), it is possible to find the output level, which maximises its profit.

Profit can be maximised by optimising the profit function, as follows;

48 FOC : f '(Q) = -2Q + 20 = 0 ⇒ Q = 10 SOC : f ''(Q) = -2 < 0

Maximum Profit = −(10) 2 + 20(10) − 5 = 95 Total cost = 4(10) 2 −12(10) +10 = 290 TotalRevenue = 3(10) 2 + 8(10) + 5 = 385

Therefore at Q=10, the firm will have the maximum Revenue.

2.2.7.5. Elasticities

In economics, price elasticity ε of demand (supply) is defined as the percentage change in quantity demanded (supplied) divided by the percentage change in price. Mathematically,

∆Q × 100 Percentage change in quantity Q ε == Percentage change in price ∆P × 100 P This is equivalent to: dQ dQ P Q dQ P εεε=⋅ or = or =⋅ Q dP dP dP Q P

Therefore, it is the shape of the demand (supply) curve that defines the elasticity of demand (supply). For example, if the demand for a commodity does not change with the changes in the price of that commodity, the demand for the commodity is said to be perfectly inelastic, figure (b) below.

The price elasticity of demand is unitless and negative, and in practice, we are interested in the of this number. Therefore, the higher the number, the more elastic is the demand curve. Following figure shows three special cases of demand curve; a) A horizontal demand curve, which is perfectly elastic, b) A vertical demand curve, which is perfectly inelastic, c) A demand curve with a negative slope, which is elastic

49

P P P

Ed>1

Ed→+∞ Ed=0 P1 Ed< 1

Ed=1 Q Q Q (a) (b) (c)

Example 1: Find the price elasticity of demand at prices, P=£100, P=£300, for the following demand function Q = 1000 − 2P i) Find the first derivative of the demand equation with respect to price

dQ d(1000 − 2P) = = −2 dP dP

ii) Find the demand when P=£100

P=£100 Q=1000-2*100=800

iii) Multiply the price-demand ratio by the value of derivative found in stage (i).

⎛100 ⎞ 1 1 Ed = (−2)⎜ ⎟ = − ⇒ (-1/4) = < 1 ⇒ inelastic demand ⎝ 800 ⎠ 4 4

ii) Now we must find the price elasticity of demand when P=300

P=£300 Q=1000-2*300=400

iii) Multiply the new price-demand ratio by the value of derivative found in stage (i).

⎛ 300 ⎞ 3 Ed = (−2)⎜ ⎟ = − then ⇒ (-3/2) = 1.5 > 1 ⇒ elastic demand ⎝ 400 ⎠ 2

Next, we need to find the quantity and price at which the elasticity is equal to one, i.e. the demand curve is unit elastic. For this purpose, we need to set the elasticity equal to one (note that it should be -1), and solve for value of Q in terms of P as follows.

dQ P P E = 1 ⇒ * = −1 ⇒ - 2* = −1 ⇒ - 2P = -Q ⇒ Q = 2P d dP Q Q

50 Substituting the value of Q in terms of P in the demand equation will result in the price at which the price elasticity of demand is equal to one.

Q = 1000 − 2P⎫ ⎬ ⇒ 2P = 1000 - 2P ⇒ 4P = 1000 ⇒ P = 250 Q = 2P ⎭

Therefore, at P=250 the price elasticity of demand is equal to one. It has is also been proved that the price elasticity of demand of a linear demand curve with a negative slope depends on the level of the price.

The summary of different forms of price elasticity of demand is presented in the table below. Elasticity of Demand Types of elasticity

Ed =0 • Perfectly Demand is not sensitive to changes in price. Inelastic 0< Ed <1 • Inelastic Demand is relatively insensitive to changes in price. Ed =1 • Unit elasticity Changes in demand are equal to changes in price. 1< Ed <∞ • Elastic Demand is relatively sensitive to the changes in price.

Ed →∞ • Perfectly Elastic Demand is sensitive to the change in price.

51

CHAPTER THREE

Differential Calculus (The multivariate case)

52 3. Differential Calculus (The Multivariate Case)

3.1. Multivariate Functions

So far we analysed functions of one variable. In real life, economic systems involve functions of more than one independent variable. For example, the demand for a commodity depends not only on the price of that commodity, but also on the price for substitute commodities as well as the income of the consumers. Therefore, it is more realistic to use functions of several variables in investigating the relationship between economic variables.

Definition: Function of two variables A function f of two variables is a rule that assigns to each ordered pair of real numbers (x,y) in a set D a unique real number denoted by f(x,y). The set D is the domain of f and its range is the set of values that f takes on, that is, {f (x, y)/(x, y)∈ D} The variables (x,y) are the independent variables and z=f(x,y). is the dependent variable.

Definition: If f is a function of two variables with domain D, then the graph of f is the set of all points (x,y,z) in ℜ3 such that z=f(x,y).and (x,y) is in D.

So the graph of a function f of two variables is a surface S with equation z=f(x,y).

3.2 Limits and continuity

Definition: Limit Let be a function of two variables whose domain D includes points arbitrarily close to (a,b). Then we say that the limit of f(x,y) as (x,y) approaches (a,b) is L and we write lim f (x, y) = L ()()x, y → a,b if for every number ε > 0 there is a corresponding number δ > 0 such that f ()x, y − L < ε whenever ()x, y ∈ D and 0 < ()()x − a 2 + y − b 2 < δ

Example 1: (The example is taken from: Stewart J; 2003; Calculus; Thomson, Brooks/Cole; p 940)

If f (x, y) = xy /(x2 + y2) , does lim f (x, y) exist? (x, y)→(0,0) Solution: If y = 0, then f (x,0) = 0 / x 2 = 0 . Therefore f (x, y) → 0 as (x, y) → (0,0) along the x-axis If x = 0, then f (0, y) = 0 / y 2 = 0 , so f (x, y) → 0 as (x, y) → (0,0) along the y-axis Although we have obtained identical limits along the axes, that does not show that the given limit is 0. Let’s now approach (0,0) along another line, say y = x. For all x ≠ 0 ,

53 x 2 1 f (x, x) → = x 2 + x 2 2 1 Therefore, f (x, y) → as (x, y) → (0,0 ) along y = x 2 Since we have obtained different limits along the different paths, the given limit does not exist. The figure below illustrates the previous example. The ridge that occurs above the line 1 y = x corresponds to the fact that f (x, y) = for all points (x,y) on that line except the origin. 2

[INSERT GRAPH 1]

Definition: Continuous function A function f of two variables is called continuous at (a,b) if lim f (x, y) = f (a,b). ()()x, y → a,b We say f is continuous on D if f is continuous at every point (a,b) in D.

Example 2: (The example is taken from: Stewart J; 2003; Calculus; Thomson, Brooks/Cole; p 943)

⎧ 3x 2 y ⎪ if (x, y) ≠ (0,0) Let f (x, y) = ⎨ x 2 + y 2 ⎪ ⎩0 if (x, y) = (0,0)

[INSERT GRAPH 2]

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We know f is continuous for (x, y) ≠ (0,0 ) since it is equal to a there. It can be shown that 3x 2 y lim f (x, y) = lim = 0 = f (0,0) (x, y)→(0,0) ( x, y)→(0,0) x 2 + y 2 Therefore, f is continuous at (0,0), and so it is continuous on ℜ2 .

Example 3: (The example is taken from: Stewart J; 2003; Calculus; Thomson, Brooks/Cole; p 943)

Where is the function h(x, y) = arctan(y/x) continuous ? Solution: The function f (x, y) = y / x is a rational function and therefore continuous except on the line x = 0. The function g(t) = arctan t is continuous everywhere. So the composite function g( f (x, y)) = arctan(x, y) = h(x, y) is continuous except where x =0. The graph below shows the break in the graph of h above the y-axis.

[INSERT GRAPH 4]

3.3 Differentiation

Let f be a function of two variables x and y, suppose we let only x vary while keeping y fixed, say y=b, where b is a constant. Then we are really considering a function of a single variable x, namely, g(x)=f(x,b). If g has a derivative at a, then we call it the of f with repect to x at (a,b) ∂f ()a,b = g ' ()a where g ()()x = f x,b ∂x By the definition of the derivative we have g(a + h)− g(a) g ' ()a = lim h→0 h and therefore

55 ∂f f (a + h,b)− f (a,b) ()a,b = lim . ∂x h→0 h

Similarly, the partial derivative of f with respect to y at (a,b) is obtained by keeping x fixed (x=a) and finding the ordinary derivative at b of the function G(y)=f(a,y)

∂f f (a,b + h)− f (a,b) ()a,b = lim ∂y h→0 h

Letting (a,b) vary in the definitions above, the partial derivatives become functions of two variables

Definition: Partial derivatives If f is a function of two variables x and y, its partial derivatives are the functions defined by

∂f f (x + h, y)− f (x, y) ()x, y = lim ∂x h→0 h ∂f f (x, y + h)− f (x, y) ()x, y = lim ∂y h→0 h

Notations for partial derivatives If z = f ()x, y , we write ∂f ∂f ∂z ()x,y = = f ()x,y = f = f = = D f = D f ∂x ∂x x x 1 ∂x 1 x

∂f ∂f ∂z ()x,y = = f ()x,y = f = f = = D f = D f ∂y ∂y y y 2 ∂y 2 y

Rules for finding partial derivatives of z = f (x, y)

1. To find fx, regard y as a constant and differentiate f(x,y) with respect to x 2. To find fy, regard x as a constant and differentiate f(x,y) with respect to y

Interpretation of partial derivatives The value of the partial derivative of f with respect to x at (x0,y0) is the slope of the curve z=f(x,y0) at the point P(x0,y0,f(x0,y0)) in the vertical plane y=y0. The tangent line to the curve at P is the line in the plane y=y0 that passes through P with this slope. The partial derivative ∂f / ∂x at (x0,y0) gives the rate of f with respect to x when y is held fixed at the value y0. The value of the partial derivative of f with respect to y at (x0,y0) is the slope of the curve z=f(x0,y) at the point P(x0,y0,f(x0,y0)) in the vertical plane x=x0. The tangent line to the curve at P is the line in the plane x = x0 that passes through P with this slope. The partial derivative ∂f / ∂y at (x0,y0) gives the rate of f with respect to y when x is held fixed at the value x0.

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[INSERT THREE FIG. 1, 2, 3] [Geometric interpretation of partial derivatives]

57 Higher derivatives

If f is a function of two variables, then its partial derivatives are also functions of two variables, so we can consider their partial derivatives, which are called second partial derivatives of f. If z = f ()x, y , we use the following notation ∂2 f ∂2 z ∂ ⎛ ∂f ⎞ = = ⎜ ⎟ = ()f xx = f xx = f11 ∂x2 ∂x2 ∂x ⎝ ∂x ⎠ ∂2 f ∂2 z ∂ ⎛ ∂f ⎞ = = ⎜ ⎟ = f = f = f 2 2 ⎜ ⎟ ()yy yy 22 ∂y ∂y ∂y ⎝ ∂y ⎠ ∂2 f ∂2 z ∂ ⎛ ∂f ⎞ = = ⎜ ⎟ = ()f yx = f yx = f21 ∂x∂y ∂x∂y ∂x ⎝ ∂y ⎠ ∂2 f ∂2 z ∂ ⎛ ∂f ⎞ = = ⎜ ⎟ = ()f xy = f xy = f12 ∂y∂x ∂y∂x ∂y ⎝ ∂x ⎠

Example 1:

Find the first second and cross partial derivatives of the following function

y = 3x 3 + 5z - 6xz 2

∂y ∂y First order partial derivatives = 9 x 2 - 6 z 2 and = 5 - 12 x z ∂x ∂z ∂ 2 y ∂ 2 y Second order partial derivatives = 18x and = -12 x ∂x 2 ∂z 2 ∂ 2 y ∂ 2 y Cross partial derivatives = -12z and = -12z ∂x∂z ∂z∂x

Theorem: Suppose f is defined on a disk D that contains the point (a,b). If the functions fxy and fyx are both continuous on D, then

f xy (a,b) = f yx (a,b)

Proposition: Tangent plane to a surface Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface z=f(x,y) at the point P(x0,y0,f(x0,y0)=z0) is

z − z0 = fx ()x0 ,y0 (x − x0 )+ f y (x0 ,y0 )(y − y0 )

Definition Linear approximation Suppose f has continuous partial derivatives. The approximation

f ()x, y = f (x0 , y0 )+ f x (x0 ,y0 )(x − x0 ) + f y (x0 ,y0 )(y − y0 ) is called linear approximation or the tangent plane approximation of f at (x0,y0).

58 Example 2: (The example is taken from: Stewart J; 2003; Calculus; Thomson, Brooks/Cole; p 960)

Find the tangent plane to the elliptic paraboloid z = 2x 2 + y 2 at the point (1,1,3).

Solution: Let f (x, y) = 2x 2 + y 2 . Then

f x (x, y) = 4x f y (x, y) = 2y

f x (1,1) = 4 f y (1,1) = 2

Then the equation of the tangent plane at (1,1,3) is

z − 3 = 4(x −1) + 2(y −1) or z = 4x + 2y − 3 The figures below show the paraboloid and its tangent plane as we zoom in toward the point (1,1,3) by restricting the domain of the function f (x, y) = 2x 2 + y 2 . Notice that the more we zoom in, the flatter the graph appears and the more it resembles its tangent plane.

[INSERT GRAPH 5]

Definition: Differentiability If z=f(x,y) then f is differentiable at (a,b) if it is possible to write

f ()()a + ∆x,b + ∆y − f a,b = f x (a,b)∆x + f y (a,b)∆y + ε1∆x + ε 2∆x where

ε1 → 0 ε 2 → 0 as (∆x,∆y) → (0,0) In other words, a differentiable function is one for which the tangent plane approximates the graph of f well near the point of tangency.

Theorem: If the partial derivatives exist near (a,b) and are continuous at (a,b), the f is differentiable at (a,b).

59 The chain rule Suppose that z=f(x,y) is a differentiable function, where x=g(t) and y=h(t) are both differentiable functions of t. Then z is a differentiable function of t and dz ∂f dx ∂f dy = + dt ∂x dt ∂y dt

Example 3: (The example is taken from: Stewart J; 2003; Calculus; Thomson, Brooks/Cole; p 968)

If z = x 2 y + 3xy 4 , where x = sin 2t and y = cos t, find dz / dt when t =0.

Solution: The Chain Rule gives dz ∂z dx ∂z dy = + dt ∂x dt ∂y dt = (2xy + 3y 4 )(2cos 2t) + (x 2 +12xy 3 )(−sin t ) It is not necessary to substitute the expressions for x and y in terms of t. We simply observe that when t = 0 we have x = sin 0 = 0 and y = cos 0 = 1. Therefore, dz = (0 + 3)(2cos0) + (0 + 0)(−sin 0) = 6 dt t=0

The derivative can be interpreted as the rate of change of z with respect to t as the point (x,y) moves along the curve C with parametric equations x = sin 2t, y = cos t. In particular, when t = 0, the point (x,y) is (0,1) and dz / dt = 6 is the rate of increase as we move along the curve C through (0,1).

Theorem: Implicit differentiation Suppose that an equation of the form F(x,y) = 0 defines y implicitly as a differentiable function of x, that is, y = f(x), where F(x,y) = 0 for all x in the domain of f. If F is differentiable we can apply the chain rule to differentiate F(x,y) = 0 with respect to x. We obtain

∂F dx ∂F dy dx + = 0 , because = 1 we have ∂x dx ∂y dx dx ∂F ∂F dy 1+ = 0 ∂x ∂y dx ∂F dy If ≠ 0 we solve for and obtain ∂y dx ∂F dy F = − ∂x = − x dx ∂F Fy ∂y

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Example 4: (The example is taken from: Stewart J; 2003; Calculus; Thomson, Brooks/Cole; p 972-973)

Find y' if x 3 + y 3 = 6xy . Solution: The given equation can be written as F(x, y) = x 3 + y 3 − 6xy = 0 So 2 2 dy Fx 3x − 6y x − 2y = − = − 2 = − 2 dx Fy 3y − 6x y − 2x

3.4 Application of Differentiation

3.4.1 Critical Values of Bivariate Functions

Definition: Local extremum A function of two variables has a local maximum at (a,b) if f (x, y) ≤ f (a,b) for all domain points (x,y) in an open disk centered in (a,b). The number f(a,b) is called a maximum value. If f ()x, y ≥ f ()a,b for all domain points (x,y) in an open disk centered in (a,b), then f(a,b) is a local minimum value.

Theorem: First Derivative Test If f has a local maximum or minimum at (a,b) and the first-order partial derivative of f exist there, then ∂f ∂f ()a,b = 0 and ()a,b = 0. ∂x ∂y

Definition: Critical point A point (a,b) is called a critical point (or ) if ∂f ∂f ()a,b = 0 and ()a,b = 0 ∂x ∂y or if one of the partial derivatives do not exist.

Remark At a critical point a function could have a local maximum, a local minimum or neither.

Definition Saddle point A differentiable function f(x,y) has a saddle point at a critical point (a,b) if in every open disk centered at (a,b) there are domain points (x,y) where f(x,y)>f(a,b) and domain points (x,y) where f(x,y)

61 Theorem: Second Derivatives test Suppose the second partial derivatives of f are continuous on a disk with center (a,b), and suppose that (a,b) is a critical point of f. Let 2 D = D()a,b = f xx (a,b) f yy (a,b) − (f xy (a,b)) Then: a) If D>0 and fxx(a,b)>0, then f(a,b) is a local minimum b) If D>0 and fxx(a,b)<0, then f(a,b) is a local maximum c) If D<0 then f(a,b) is a saddle point d) If D=0 the test is inconclusive

In summary

TURNING First Order Necessary *Second Order Sufficient Condition POINT TO BE Condition 2 2 Maximum ∂y ∂y ∂ y ∂ y 2 2 2 2 = 0 and = 0 < 0 and < 0 ∂ y ∂ y ⎛ ∂ y ⎞ 2 2 ⋅ > ⎜ ⎟ ∂x ∂z ∂x ∂z 2 2 ⎜ ⎟ ∂x ∂z ⎝ ∂z∂x ⎠ Minimum ∂y ∂y 2 2 2 2 2 2 = 0 and = 0 ∂ y ∂ y ∂ y ∂ y ⎛ ∂ y ⎞ 2 > 0 and 2 > 0 ⋅ > ⎜ ⎟ ∂x ∂z ∂x ∂z 2 2 ⎜ ⎟ ∂x ∂z ⎝ ∂z∂x ⎠ 2 2 2 Saddle Point ∂y ∂y ∂ y ∂ y ∂ 2 y ∂ 2 y ⎛ ∂ 2 y ⎞ = 0 and = 0 > 0 , < 0 ⋅ < ⎜ ⎟ 2 2 2 2 ⎜ ⎟ ∂x ∂z ∂x ∂z ∂x ∂z ⎝ ∂z∂x ⎠ or vice versa * Applies only if the first order necessary condition is met

Example 1: A firm producing two goods x and z has the following profit function

Π = 64x − 2x 2 + 4xz − 4z 2 + 32z −14

Find the profit maximising level of output for each of the goods, that is find the combination of x and z that gives maximum profit. i) Take the first order partial derivatives and set them equal to zero. Then solve for x and z simultaneously.

∂Π FOC : = 0 ⇒ 64 - 4x + 4z = 0 ∂x ∂Π = 0 ⇒ 4x -8z + 32 = 0 ∂z ⎧- 4x + 4z = -64 Solve the simultaneous equations ⎨ ⇒ z = 24 x = 40 ⎩ 4x -8z = -32

Now we should check that this turning point, A(x=40,z=24) is a maximum or minimum. ∂ 2Π ∂ 2Π SOC : = - 4 < 0 = -8 < 0 ∂x 2 ∂z 2

62 Since both the second partial derivatives are less than zero, the turning point might be a maximum. In order to ensure the existence of this maximum, we should take the cross partial derivatives and test whether the product of the second partial derivatives is greater than the product of the cross partial derivatives.

∂ 2Π ∂ 2Π Cross partial derivatives : = 4 = 4 ∂x∂z ∂z∂x Therefore 2 ∂ 2Π ∂ 2Π ⎛ ∂ 2Π ⎞ > ⎜ ⎟ ⇒ (-8)(-4) > (4)2 2 2 ⎜ ⎟ ∂x ∂z ⎝ ∂x∂z ⎠

This means that there is a maximum point A(x=40,z=24). Now the task is to find the maximum value of profit, that is to substitute the values of x=40 and z=24 in the profit function. P=64xx - 222 + 4 xzz - 4 + 32 z - 14

p=-+64 ()( 40 2 4022 ) 4 ()()( 40 24 -+-= 4 24 ) 32 () 24 14 1650

Example 2: A firm's total costs are related to its work force (L) and capital equipment (K) by the function, TC = 10 L2 + 10 K 2 - 25 L - 50 K - 5 L K + 2000 . Find the combination of K, L to minimise TC. ∂ (TC) FOC: = 20 L - 25 - 5 K = 0 (1) ∂L ∂ (TC) = 20 K - 50 - 5 L = 0 (2) ∂K This is a system of two equations and two unknowns, K and L.

4 x (1) + (2) gives: 75 L - 150 = 0, ⇒ L = 2 L = 2 ⇒ K = 3. To verify it is a minimum take the SOC's SOC: ∂ 2 (TC) ∂ 2 (TC) = 20 > 0 and = 20 > 0 ∂L2 ∂K 2

Now test the existence using the product of the cross partial derivatives:

2 ∂ 2 (TC) ∂ 2 (TC) ⎛ ∂ 2 (TC) ⎞ ≡ = - 5 ⇒ ⎜ ⎟ = 25 ∂K ∂L ∂L∂K ⎝ ∂K ∂L ⎠

∂ 2 (TC) ∂ 2 (TC) × = 400 > 25 ∂L2 ∂K 2

∴ There is a turning point at K = 3, L = 2, and it is a minimum. At the above point, TC is: TC = 10 × 22 + 10 × 32 - 25 × 2 - 50 × 3 - 5 × 2 × 3 + 2000 = 1900

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3.4.2. Critical Points of Multivariate Functions

Let us consider a function y=f(x) = f(x1, x2, …, xN). We define the vector (or simply gradient) as

⎛∂y / ∂x ⎞ ⎜ 1 ⎟ ∂f (x) ⎜∂y / ∂x2 ⎟ = ⎜ ⎟ ∂x M ⎜ ⎟ ⎜∂y / ∂x ⎟ ⎝ N ⎠

Similarly, we define the Hessian as the matrix containing the second partial derivatives and cross partial derivatives; it is a square symmetric matrix with the following form

⎛∂ 2 y / ∂x ∂x ∂ 2 y / ∂x ∂x ∂ 2 y / ∂x ∂x ⎞ ⎜ 1 1 1 2 L 1 N ⎟ ∂ 2 f (x) ⎜∂ 2 y / ∂x ∂x ∂ 2 y / ∂x ∂x ∂ 2 y / ∂x ∂x ⎟ H = = 2 1 2 2 L 2 N ⎜ ⎟ ∂x∂x' ⎜ M O M ⎟

⎜ 2 2 2 ⎟ ⎝∂ y / ∂xN ∂x1 ∂ y / ∂xN ∂x2 L ∂ y / ∂xN ∂xN ⎠

The necessary and sufficient conditions for the maximisation (minimisation) problem in the multivariate case are generalization of the two variable case.

First order condition

For a general function of n variables, say y = f (x1 ,..., x n ), the FOC's for a turning point amounts to: ∂y ∂y ∂y = = K = = 0 ∂x1 ∂x 2 ∂x n

Second order conditions Let the Hessian be ⎛ f f f ⎞ ⎜ 11 12 K 1n ⎟ ⎜f 21 f 22 K f 2n ⎟ Η = ⎜ ⎟ ⎜ M M O M ⎟ ⎜ ⎟ ⎝f n1 f n2 L f nn ⎠

∂ 2 y ∂ 2 y ∂ 2 y ∂ 2 y where f11 = 2 , f12 = , f 22 = 2 , f 21 = , etc ∂x1 ∂x1∂x 2 ∂x 2 ∂x 2∂x1

Denote successive principal minors of H by

f11 f12 Η1 = f11 , Η 2 = , K , Η n = Η f 21 f 22

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Then, the SOCs are as follows: • for a maximum, H must be negative definite. Equivalently:

, H1 < 0, H 2 > 0, H 3 < 0, K • for a minimum H must be positive definite; that is:

H1 > 0, H 2 > 0, H 3 > 0, K

a) Plot of a multivariate function b) Maximum and minimum

Example 1: 2 2 2 Consider the function y = x1 - 2x1x 2 + 2x 2 + 2x1x 3 + 4x 3 - 2x 3

FOC: ∂y/∂x1 = 2x1 - 2x 2 + 2x 3 = 0

∂y/∂x 2 = - 2 x1 + 4 x 2 = 0

∂y/∂x 3 = 2 x1 + 8 x 3 - 2 = 0

When solved simultaneously, the system gives x1 = -1, x2 = -0.5, x3 = 0.5. That is, there is a turning point at (-1, -0.5, 0.5).

SOC: When all the 2nd order and cross partial derivatives of the function are derived and evaluated at the turning point, the following (3x3) describes the curvature of the function at the turning point. ⎛ 2 − 2 2⎞ ⎜ ⎟ Η = ⎜− 2 4 0⎟ ⎜ ⎟ ⎝ 2 0 8⎠

The values of its successive principal minors are: 2 − 2 Η = 2, Η = = 4, Η = 16 1 2 − 2 4

Since they are all positive, i.e. the Hessian is positive definite, thus the turning point is a minimum.

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Example 2: A firm uses three types of advertisements, (TV commercials, Radio commercials and magazine advertisements). The costs of these advertisements are as follows; x1, x2 and x3 (in million pound) for TV, Radio and Magazine ads, respectively. The firms total revenue can be explained as a function of these advertisements as follows;

2 2 2 TR = −5x1 +10x1 + x1 x3 − 2x2 + 4x2 + 2x2 x3 − 4x3

Find the optimum amount of the expenditure on these commercials, which maximises the firm’s total revenue.

First and Second Partial derivatives: ∂y ∂y ∂y = y1 = −10x1 +10 + x3 = y2 = - 4x 2 + 2x3 + 4 = y3 = x1 + 2x2 −8x3 ∂x1 ∂x2 ∂x3

y11 = −10 y12 = 0 y13 =1

y21 = 0 y22 = −4 y23 = 2

y31 =1 y32 = 2 y33 = −8

FOC: Setting the first partial derivative equal to zero and solving the system of 3 equations and three unknowns will result in

⎧10x1 − x3 =10 ⎪ ⎨ 4x2 − 2x3 = 4 ⇒ x1 ≅ 1.04 , x 2 ≅ 1.22 , x 3 ≅ 0.43 ⎪ ⎩x1 + 2x2 −8x3 = 0

SOC: Construct the Hessian using the second partial derivatives and calculate the determinants of principle minors

−10 0 1 H = 0 - 4 2 1 2 -8 −10 0 1 -10 0 H = −10 < 0 , H = = 40 > 0 , H = 0 - 4 2 = -276 < 0 1 2 0 - 4 3 1 2 -8

Determinants of principle minors alternate sign, therefore the turning point is a maximum; that is the revenue of the firm would be maximum if the expenditure of advertisements is set to be x1 ≅ £1.04m , x 2 ≅ £1.22m and x 3 ≅ £0.43m , for TV, Radio and magazines, respectively. Maximum revenue would be?

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3.4.3. Constrained Optimisation

Since resources are limited in everyday life, we often have to take decisions of the type: 1. The firm wants to maximise revenue subject to costs. 2.The household wants to maximise utility subject to the budget constraint.

Therefore, we need to consider such restrictions in our optimisation process.

This means optimising an objective function, f(x), subject to certain limitations or constraints, let the constraint be a function itself, g(x).

For simple functions it may be possible to substitute the constraint into the objective function and optimise this new function in the usual way.

3.4.3.1.Method of substitution

When the constraint function is simple; that is linear and bivariate (az+bx=c). It is convenient to find the one variable with respect to the other one and substitute the result in the objective equation and perform optimisation.

Example 1: Minimise y = 1 + 2 x + 5 z2 subject to x = z. Substitute the constraint in the objective function to eliminate x,y =1 + 2z + 5z2 . Minimisation yieldsdy/dz = 2 + 10z = 0 . Whence z= -1/5, and as a result x= -1/5.

3.4.3.2.

In many cases the optimisation is done with respect to more than two or three variables or the constraint might be nonlinear. Therefore, it is not always easy to use the substitution method. An alternative is to use the Lagrange Multiplier method. This method adds a multiple of the constraint function to the objective function and optimises the new function know as Lagrangean. Another advantage of the Lagrange Multiplier method is that it provides additional information regarding the economic relationship between the constraint and the objective function.

Let y be a function of x and z, y=f(x,z). Let’s also assume we want to optimise subject to the following constraint: g(x,z)=k. In general, to find the solution to a constrained optimisation problem we want to optimise a new function called the Lagrangean:

L(x,z,λ) = f(x,z) + λ[k − g(x,z)], where λ is some constant to be determined. Note that the term in the square bracket is equal to zero by construction, therefore, the Lagrangean function only restrict the values that x and z can take but do not affect f(x,z).

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Example 1: Suppose that a company produces two different types of goods, x and z, and has the following total cost function;

TC = 8x 2 − xz +12z 2

The company is also bound by the existence of the raw material to produce a total number of 42 units. Find the values of x and z that optimises the cost of the company, subject to the constraint using Lagrange Multiplier method.

In order to optimise the cost function subject to the constraint, first rearrange the constrain function to get zero in one side.

x + z = 42 ⇒ 42 - x - z = 0 then construct the lagrangean function using the constrain function;

L = f (x, z) + λ[k − g(x)] = 8x 2 − xz +12z 2 + λ[42 − x − z]

L = 8x 2 − xz +12z 2 + 42λ − λx − λz Take the first partial derivatives of L with respect to x,z and λ. ∂L =16x − z − λ = 0 ∂x ∂L Solve the system of 3 variable 3 equations. = 24z − x − λ = 0 ∂z ∂L = 42 − x − z = 0 ∂λ

⎧λ =16x − z ⎧ 25z ⎪ ⎧16x - z = 24z - x ⎧17x = 25z ⎪x = ⎨λ = 24z − x ⇒ ⎨ ⇒ ⎨ ⇒ ⎨ 17 ⎪ ⎩42 - x - z = 0 ⎩42 - x - z = 0 ⎪ ⎩42 − x − z = 0 ⎩42 − x − z = 0

⎧z =17 ⎧ 25z ⎪ ⎨42 − − z = 0 ⇒ z =17 ⇒ ⎨x = 25 ⎩ 17 ⎪ ⎩λ = 383

If the production held at x=25 and z=17, the total cost of the company is minimised while the constraint is also met. That is x+z=17+25=42.

Example 2: The production function of a firm describes the technological relationship between inputs (raw materials, labour etc) and output. Assume, a particular firm wants to optimise a Cobb- Douglas production function, Q = 10 L1/2 K1/2 , subject to the cost of input, 4 L + 10 K = 100 , where L and K are Labour and Capital inputs.

68 Rearrange the constraint to get:100 - 4 L - 10 K = 0 , and build the Lagrangean:

L* = 10 L1/2 K1/2 + λ (100 - 4 L - 10 K).

We want to maximise this with respect to L, K and λ. The FOC yields:

∂L* 5K1/2 ≡5 L-1/2 K1/2 - 4 λ = 0 ⇒ λ = ∂L 4L1/ 2 ∂L* 5L1/2 ≡ 5 L1/2 K -1/2 - 10 λ = 0 ⇒ λ = ∂K 10K1/2 ∂L* ≡ 100 - 4 L - 10 K = 0 ⇒ 100 - 4 L - 10 K = 0 ∂λ

This is a system of 3 equations in 3 unknowns, L, K and λ. Hence 5K1/ 2 5L1/ 2 5 = ∴50 K = 20 L ∴ L = K 4L1/ 2 10K1/ 2 2

100 - 4 L -10 K = 0 ∴ 100 - 4L - 10K = 0 ∴ 10K + 10K = 100 therefore, K = 5 and L = 12.5 maximise production. Also λ = 0.79. The number of production units at the maximum is (K=5, L=12.5) =79.06

The Lagrange multiplier, λ, is also known as the shadow price. It shows by how much the optimum Q would change when the constraint is relaxed by 1 unit. In the above example λ = 0.79. Thus, if the constraint is relaxed from 100 to 101 the optimal level of output would increase by 0.79 units to Q* = 79.85.

Note: When λ = 0 the constraint is not binding; it does not affect the solution.

3.4.3.3.Bordered Hessian

Just as with unconstrained optimization problems, the SOC's need to be considered in constrained optimization problems to determine the nature of the turning point of the Lagrangean function.

Assuming f(x) is the objective function and g(x) is the constraint, the following Bordered Hessian matrix determines the curvature of the Lagrangean function.

⎛ 0 g g g ⎞ ⎜ 1 2 K n ⎟ ⎜ g1 f11 + λ g11 f12 + λ g12 K f1n + λ g1n ⎟ HB = ⎜ g f + λ g f + λ g f + λ g ⎟ ⎜ 2 21 21 22 22 K 2n 2n ⎟ ⎜ M M M O M ⎟ ⎜ ⎟ ⎝ g n f nl + λ g nl f n2 + λ g n2 K f nn + λ g nn ⎠

69 At the turning point (when the FOC's are satisfied), there is a: maximum when successive principal minors of HB greater than second order alternate in sign, starting from a positive value: HB3 > 0, HB4 < 0, HB5 > 0, K minimum when successive principal minors of HB are negative

HB3 < 0, HB4 < 0, HB5 < 0, K

It is also possible to have several constraints in a constrained optimisation problem. In this case the Lagrangean function may be written as

L( x, λ1 ,λ2 ,...,λn ) = f(x) + [ λ1g1 (x) + λ2g 2 (x) + K + λn g n (x) ]

These are complex mathematical programming problems, which may be solved using specialised computer software.

• Consider the Bordered Hessian for Example 1

0 1 1

HB = 1 16 -1 ⇒ HB3 = −42 < 0 therefore the turning point is a minimum. 1 -1 24

70

CHAPTER FOUR

Integral Calculus

71 4 Integral Calculus

4.1. Integration

4.1.1. Definite and Indefinite Integrals

Consider the graph below, where the curve represents a function f(x). The shaded portion of the figure consists of those points, which lie below the curve C and above the horizontal axis.

1. Divide the range of values of x into n subinterval, each of size ∆x.

2. Calculate the value of the function at the end-point of each subinterval, say

f(x1 ), f(x2 ), K ,f(xn ).

3. Sum the of the rectangles n Sn = Σ f(xi ) ∆x i=1

If we let ∆x → 0 , or equivalently n → ∞ , the sum of the area of the rectangles tends to the area under the curve n A = lim Σ f(xi ) ∆x n→∞ i=1

y

f(x1) f(x2)

f(x3)

f(x4)

∆x1 ∆x2 ∆x3 ∆x4 X

This limit is known as the integral of f(x) between values a and b, and is written symbolically as b A = ∫ f(x) dx . a

More rigorously, an integral is defined as follows:

72

Definition: Definite Integral * Let f(x) be defined on [a,b] and let x0, x1, … ,xn be a partition of [a,b]. For each [xi-1,xi], let xi ∈ [xi-1, xi]. Then we define the (definite) integral of f between a and b as:

b n * f (x)dx = lim f (xi )∆xi ∫ max ∆x→0 ∑ a i=1

Definition: Primitive (anti-derivative) of a function We now define a primitive function (or anti-derivative) F of a function f to be any function for which dF()x F '()x = f ()x , i.e. = f ()x dx The symbol ∫ f ()x dx is used to denote a general primitive function of f. In this notation an equation like ∫ f ()x dx = F ()x + c, where c is an arbitrary constant, is considered to be an alternative way of writing

F , ()x = f ()x . b Despite similarity in appearance, the symbols ∫ f (x)dx and ∫ f ()x dx represent different a concepts. However, the two processes are related by the Fundamental Theorem of Calculus.

Theorem: If F (x) is the a primitive of f(x) then

b x=b ∫ f(x) dx = [ F(x) ] x=a = F(b) - F(a). a

Observation: Very often the symbol ∫ f ()x dx is referred to as indefinite integral rather than as a primitive b or . To distinguish the symbol ∫ f (x)dx from ∫ f ()x dx , the latter is called a a definite integral.

73

Example: 2 3 4 2 4 4 ∫ x dx = [ x / 4 ] 0 = F( 2 ) - F(0 ) = ( 2 / 4 ) - (0 / 4 ) = 4 0

10

8 6

4 2

0 -3 -2 -1 0 1 2 3 -2

-4 -6

-8 -10

y0=x^3

4.1.2. Properties of definite integrals

b a a 1) ∫ f(x) dx = - ∫ f(x) dx 2) ∫ f(x) dx = 0 a b a

If f(x) is continuous in the domain [a, b], and there is point c between a and b where a < c < b , then we can write

c b b 3) ∫ f(x) dx + ∫ f(x) dx = ∫ f(x) dx a c a

b b 4) ∫ k f(x) dx = k ∫ f(x) dx a a

b b b 5) ∫ [ f(x) ± g(x) ] dx = ∫ f(x) dx ± ∫ g(x) dx a a a

What happens if f(x) has negative values in interval [a, b]

We assumed that f(x) was positive in interval [a, b]. This is because the area should always be positive. Now if we find that at some intervals [a, c] (a

74

Example: Find the area between the curve and the x-axis for the function y=x for the interval [-1, 1].

1 1 ⎡ x 2 ⎤ ⎡(1) 2 (−1) 2 ⎤ A = xdx = ⎢ ⎥ = ⎢ − ⎥ = 0 ∫ 2 2 2 −1 ⎣ ⎦ −1 ⎣ ⎦

-1 1

It can be seen that if we take the finite integral, the result would be zero. In order to find the area we need to use the above mentioned properties of definite integrals. Therefore we can split the interval into two sub-intervals [-1, 0] and [0, 1].

0 1 0 1 ⎡ x 2 ⎤ ⎡ x 2 ⎤ ⎡(0) 2 (−1)2 ⎤ ⎡(1) 2 (0)2 ⎤ A = ∫ xdx + ∫ xdx = ⎢ ⎥ + ⎢ ⎥ = −⎢ − ⎥ + ⎢ − ⎥ = 1 −1 0 ⎣ 2 ⎦ −1 ⎣ 2 ⎦ 0 ⎣ 2 2 ⎦ ⎣ 2 2 ⎦

4.1.3. Rules of Integration - Indefinite Integral

We saw that a set of rules need to be remembered to perform differentiation. Similarly, a set of rules must be known to perform integration. Here are the most important ones.

Note that you need to check the result by differentiation of your answer.

1) The integral of a constant is a linear function:

∫ a dx = a x + c a, c∈R

Examples:

∫ −2 dx = -2x + c Check: y = −2x + c ⇒ y' = − 2

∫10 dx = 10x + c Check: y =10x + c ⇒ y' = 10

75

n 2) The integral of a power function, x , where n≠-1, is :

1 ∫ x n dx = x n+1 + c for n ≠ -1 n + l Examples: 1 1 a) ∫ x−2 dx = x-2+1 + c = -x-1 + c = - + c -2 +1 x Check: y = −x −1 + c ⇒ y' = −( −1)x −1−1 = x−2

1 10 b) ∫10 x5dx = 10* x5+1 + c = x6 + c 5 +1 6 10 10 Check: y = x6 + c ⇒ y' = ( 6x5 ) =10x5 6 6

3) The integral of a constant times a function is equal to the constant times the integral of that function, therefore:

∫ k ⋅f(x)dx = k ∫ f(x)dx

Example: 1 a) 4x 2dx = 4 x2dx = 4* x3 + c ∫ ∫ 3

4) The integral of a function raised into a power, is:

1 ∫ (ax + b)n dx = (ax + b)n+1 + c a(n +1)

Example: 1 a) ∫ ( 2x + 4 )3dx = ( 2x + 4 )3+1 + c 2( 3+1) 1 b) ∫ (x + 4 )1/ 2dx = (x + 4 )(1/ 2 )+1 + c 1((1/ 2 )+1)

5) The integral of an exponential function, is:

f(x) f(x) ∫ f '(x) e dx = e + c

where a special case is:

∫ex dx = ex + c

76 Example: 2 2 a) ∫ 2x ex dx = e x + c

3 3 b) ∫ 3x2 e x +1 dx = e x +1 + c

6) The integral of inverse of x is equal to the natural log of x plus a constant;

1 ∫ dx = lnx + c x

Where a x>0.

7) The integral of the derivative of a function times the inverse of the function is equal to the natural log of that function plus a constant:

f'(x) dx = ln(f(x)) + c ∫ f(x)

Example:Where a f(x)>0.

( 2x +1) a) dx = ln(x2 + x) + c ∫ (x 2 + x)

( 3x 2 + 2x + 4 ) b) dx = ln(x3 + x 2 + 4x) + c ∫ (x3 + x 2 + 4x)

8) The integral of sum or differences of two functions is equal to the sum or differences of the integrals of those two functions.

∫[f(x) ± g(x)] dx = ∫f(x)dx ± ∫g(x)dx

Example: a) ∫ [(x +1)+(x2-4 )] dx = ∫∫(x +1)dx + (x 2-4 )dx x 2 x3 = ( + x)+( − 4x)+ c 2 3

77

9) Integral by parts

∫f(x)g'(x)dx = f(x)g(x) - ∫f'(x)g(x)dx The equation expresses a relationship which can be used to determine integrals where the integrand has the form f(x)g'(x). This is derived from the product rule in differentiation.

Example: a) 2 ∫ x ln x dx

let us assume f(x)=ln(x) and g'(x) =x2

1 x3 Then we can write f'(x) = and g(x) = x 2dx = x ∫ 3

Substituting into the original integral yields

3 3 2 ⎛ x ⎞ x 1 ∫ x ln x d(x) = ( ln x)⎜ ⎟ − ∫ ⋅ dx ⎝ 3 ⎠ 3 x

x3 x 2 x3 x3 = ln x − ∫ dx = ln x − + c 3 3 3 9

10) Integral by partial fractions

Often rational functions cannot be integrated directly. In such cases one

may be able to restate the rational function in an equivalent form

consisting of more elementary functions.

Example: x + 3 a) dx ∫ x 2 + 3x + 2

the above integral can be written as

x + 3 2 1 dx = dx − dx ∫ x2 + 3x + 2 ∫ x +1 ∫ x + 2

= 2ln(x +1)- ln(x + 2 ) + c

78 x3 − 2x b) dx ∫ x −1

using division rules for mathematical expressions we can write

x3 − 2x ⎛ −1 ⎞ dx = ⎜ x2 + x −1+ ⎟dx ∫ x −1 ∫ ⎝ x −1⎠

1 = ∫ ()x2 + x −1 dx − ∫ dx x −1 x3 x2 = + − x − ln(x −1) + c 3 2 Example:

Figure 6.1 plots three different functions, namely; y0 = x3 , y1 = x3 + 2 and 6 3 y2 = x -2. These functions are all obtained 4 2 through, ∫3x dx , however the value of c 2 varies (0, 2 and –2). 0 -4-2024 Therefore, we need more information to -2 obtain exactly the same primitive function through integration. The extra information is -4 usually given by a point, which satisfies the -6 equation. y0=x^3 y1=x^3+2 y2=x^3-2

4.1.4. Auxiliary Conditions

An auxiliary condition is required to find the value of the constant in the primitive function. If it is known that the solution passes through point (a,b) we can substitute the values of x and y in the primitive function and solve for c. For instance, if we are also told in the above example that when x=1, y=3, substituting in the above and solving for c would provide its value. That is, substitution in y = x4 + c yields 3 =14 + c, ∴ c = 2.

4.2. Applications of Integrals

4.2.1. Deriving Totals (Revenue/Cost) from Marginals (Revenue/Cost)

MR is the derivative of the total revenue function (TR). Therefore, the anti-derivative of the MR function gives the TR function.

Similarly, MC is the derivative of the total cost function (TC). Therefore, the anti-derivative of the MC function gives the TC function.

79

Let MC = 100 + x TC = ∫ MC dx = ∫ (100 + x) dx = 100 x + (1/2) x 2 + c

More information is needed to determine c in the TC function. Thus, if TC = 40,000, at x = 100, then 40,000 = 100 (100) + (1/2) 1002 + c ∴ c = 40,000 - 10,000 - 5,000 = 25,000 Hence: TC = (1/2) x 2 + 100 x + 25,000

4.2.2. Present Value of Cashflows

We can also use integrals to discount a continuous stream of income, f(t), at a specified interest rate, r, received during time interval [0,T].

T PV = ∫ f (t)e −rt dt t=0

Example 1: Find the PV of a continuous income stream flowing at a rate of £1000 per year over the next 10 years, assuming an interest rate of r=8%=0.08 annually, compounded continuously.

10 T 10 −0.08t −rt −0.08t ⎡ ⎛ e ⎞⎤ PV = f (t)e dt = 1000e dt = ⎢1000⎜− ⎟⎥ ∫ ∫ ⎜ 0.08 ⎟ t=0 t=0 ⎣ ⎝ ⎠⎦ 0 1000 = (1− e −(10)(0.08) ) ≈ 6883.39 0.08

Example 2: Find the PV of a perpetual income stream flowing at the uniform rate of a dollars per year, when t→∞, assuming an interest rate, r, compounded continuously.

∞ ∞ −rt ∞ ⎛ ⎞ −rt ⎡ ⎛ e ⎞⎤ ⎡ a −rt ⎤ ⎡ a −rt ⎤ ⎡ a −rt ⎤ PV = ae dt = ⎢a⎜− ⎟⎥ = − ()e = ⎜ − ()e − − ()e ⎟ ∫ ⎜ r ⎟ ⎢ r ⎥ ⎜ ⎢ r ⎥ ⎢ r ⎥ ⎟ t=0 ⎣ ⎝ ⎠⎦ 0 ⎣ ⎦ 0 ⎝ ⎣ ⎦ t→∞ ⎣ ⎦ 0 ⎠

⎛ ⎡ a ⎛ 1 ⎞⎤ ⎡ a ⎛ 1 ⎞⎤ ⎞ ⎡ a ⎛1⎞⎤ a ⎜ − ⎜ ⎟ − − ⎜ ⎟ ⎟ = 0 − − ⎜ ⎟ = ⎜ ⎢ r e rt ⎥ ⎢ r e rt ⎥ ⎟ ⎢ r 1 ⎥ r ⎝ ⎣ ⎝ ⎠⎦ t→∞ ⎣ ⎝ ⎠⎦ 0 ⎠ ⎣ ⎝ ⎠⎦

80 4.2.3. Measuring Probabilities

Integrals are used in statistics when calculating probabilities from continuous distributions, as the area under the density function curves. Thus, b P( a ≤ x ≤ b ) = ∫ f(x) dx a

0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 -4 -2 0 2 4

normal dist

The area under the standardised curve is equal to one, which represents 100% probability. The area under this curve between [–1.96 and 1.96] is 0.95, which represent 95% probability.

4.3 Multiple Integration

4.3.1. Double Integral of a Function of Two Variables

Let f(x,y) be defined on a closed and bounded region R of the xy-plane. Set up a grid of vertical and horizontal lines in the xy-plane to form an inner partition of R into n rectangular sub-regions Rk of area ∆Ak, each of which lies entirely in R. (Ignore the rectangles that are not * * entirely contained in R) Choose a point (xk , yk ) in each sub-region Rk. The sum

n * * Sn = ∑ f (xk , yk )∆Ak k=1

This is called a . In the limit, as we make our grid more and more dense, we define the double integral of f(x,y) over R as

n f(x,y)dA = lim f (x* , y* )∆A ∫∫ ∑ k k k max ∆AK →0 R k=1

Notes • If this limit exists, we say that f is integrable over the region of integration R. • If f(x,y) is continuous on R, then f(x,y) is integrable over R.

81

4.3.2 Geometric Interpretation of the Double Integral

n Notice that as we increase the density of our grid, the sum ∑ Ak of the individual rectangles k=1 better and better approximates the area of region R. In the limit as ∆Ak → 0, we have

Area of R = ∫∫dA R

* * Suppose now that f(x,y) ≥ 0 on R. Then f(xk ,yk )∆Ak is the volume of a rectangular * * parallelepiped of height f(xk ,yk ) and base area ∆Ak. Adding up these volumes, we get an approximation for the volume of the solid above R and below the surface z = f(x,y). Thus, in the limit as ∆Ak → 0,

Volume of solid above R ⎫ ⎪ and below the surface ⎬ = ∫∫ f(x,y)dA (for f(x,y) ≥ 0 on R) ⎪ R z = f(x,y) ⎭

Note : The interpretation of the double integral as a volume still holds if f(x,y) takes on both positive and negative values. In this case, we obtain the difference between the volume above the xy-plane between z = f(x,y) and R and the volume below the xy-plane between z = f(x,y) and R.

Example Let's evaluate the double integral ∫∫6xydA where R is the region bounded by y = 0, x = 2, R and y = x2.

2 x2 ∫∫6xydA = ∫ ∫ 6xy dydx R 00 2 2 2 x2 1 1 1 = [3xy 2 | ]dx = 3x 5 dx = x 6 = ( 64 ) − ( 0 ) = 32 ∫ y=0 ∫ 2 2 2 0 0 x=0

82

CHAPTER FIVE

Ordinary Difference and Differential Equations

83 5. Difference and Differential Equations

5.1. Basic Idea of a Difference Equation

A difference equation expresses the value of an endogenous variable in a given period as a function of the same variable in past periods, together with any exogenous variables. For example,

YfYYttttn= ()−−12, ,....., YX − , (1)

This states the value of Y in period t depends on the values of Y in earlier periods

(Yt-1,....., Yt-n) and the exogenous variable X.

5.2. Linear Difference Equations

The most common type of difference equation encountered in economic models is the linear difference equation. This often arises as the reduced form equation of a dynamic linear economic model (as the examples below will show). In general a linear difference equation of order n (with constant coefficients and constant term) is written as follows:

(2) YaYaYtt=+11−− 2 t 2+.....+ aYb ntn −+ where a1, ... , an are given coefficients/parameters and b is a given constant. Thus a first-order linear difference equation would be written in general as

YaYbtt=+11− (3) and a second-order equation would be

(4) YaYaYbtt=+11−− 2 t 2+

• Note that if the constant, b, is zero then the equation is called homogeneous. If b ≠ 0 the equation is non-homogeneous.

84 5.3. Numerical Example

Consider the following first-order equation:

YYtt=+08. −1 30 (5)

This says that the value of Y in any period is equal to the value of Y in the previous period plus 30. So clearly the value of Y changes over time such that the values in each successive period are linked as shown by the difference equation. Given the value of Y in some particular period (e.g. period 0) we can use the difference equation to successively find the value of Y in each subsequent period.

Thus, suppose that Y is equal to 10 in period 0, i.e. Y0 = 10. Then,

YY10=+=+=08.. 30 08( 10) 30 38

YY21=+=+=08.. 30 08() 38 30 60 . 4

YY32=+=08.... 30 08() 604 += 30 7832 and so on.

• Clearly, the problem with using this method to find the value of Y in any given period is that we have to calculate Y in all the periods prior to the one we are interested in. It would be much better if we had a formula which told us the value of Y in any given period t. This is the general solution to a difference equation which we discuss below.

5.4. Solution of a First-Order Linear Difference Equation

By the general solution to a difference equation we mean a function of time t which gives the value of Y in any period. Before looking at this let’s determine the equilibrium value of Y.

5.4.1. Equilibrium Solution

Given the first-order, linear, non-homogeneous difference equation

YaYbtt=+−1 (6) the equilibrium solution is the value of Y that, once attained, will remain constant throughout time. If we call this equilibrium value Y* then in equilibrium we have,

∗ (7) YYtt==−1 Y

85

Hence, by substitution into the first-order equation we get:

(8) YaYb∗∗=+ and solving for Y* gives

YaYb∗∗−= Yab∗ ()1−= b Y ∗ = (9) ()1− a

5.4.2. General Solution

The general solution to a first-order, linear, non-homogeneous difference equation can be shown to be

t ∗ (10) YKaYt =+

This gives the value of Y in any period t as a function of t. Thus, to find the value of Yt we simply substitute into this general solution the value of t.

Note, however, that the general solution contains a constant K whose value we do not know. To determine its value we need to specify the initial condition - i.e. the value of Y in period 0.

Assume that in period 0 the value of Y is Y0. This implies that,

0 ∗∗ (11) YKaY0 =+=+ KY

Hence, ∗ KY=− Y 0 (11')

Substituting this expression for K into the general solution gives,

t ∗∗ (12) YaYYt =−+()0 Y

This is the general solution we can use to solve any first-order, linear, non-homogeneous difference equation.

86

Numerical Example Continuing with our example equation

YYtt=+08. −1 30 (13) we know that the equilibrium value Y* = 150. Suppose that Y0 = 10, then the general solution is given by t Yt =−+08. () 10 150 150 or, Y =−140 08. t + 150 t ()

We can now easily find the value of Y in any period by substituting into this general solution the appropriate value of t. So, for example, if t = 3 then

3 Y3 =−140() 08. + 150 =−140() 0512. + 150 = 78. 32

• Note that this confirms the value of Y3 we calculated earlier using the method of

calculating successive values of Y starting from Y0. The main advantage of using the

general solution is that we can immediately calculate the value of Yt we are interested in without having to calculate all the prior ones

5.4.3. Stability Analysis

An equilibrium solution/value is stable if the value of Y converges onto the equilibrium value as t gets larger (i.e. as t approaches infinity). Stability of the equilibrium depends on the value of a. Specifically, there are four possibilities:

1. If a > 1 then the equilibrium is unstable. Y will grow continuously since at becomes larger as t → ∞ 2. If 0 < a < 1 the equilibrium is stable with a smooth convergence to equilibrium 3. If -1 < a < 0 the equilibrium is stable but with a declining (‘damped’) cyclical convergence to equilibrium 4. If a < -1 the equilibrium is unstable with increasing (‘explosive’) cycles.

So, for stability a must be less than one in absolute value (i.e. ignoring the sign) −11< a <

87

Numerical Examples

In our example equation

YYtt= 08. −1 + 30 the equilibrium value of 150 will be stable because a = 0.8. To confirm that the value of Y will indeed converge on to 150 as t increases we can calculate Y for successive values of t

(assuming Y0 = 10) and plot the graph of Yt against t. The result is shown below:

160 150 140 130 120 110 100 90

Y 80 70 60 50 40 30 20 10 0 0 1020304050 t

Note that since a is between zero and one the convergence to the equilibrium value is smooth.

Let’s consider an example now of damped cyclical convergence. As indicated above this occurs if a is less than zero but larger than -1. Thus, suppose we have the following difference equation:

YYtt= −0. 9−1 + 114

Here a = - 0.9 and b = 114. Hence the equilibrium value is b 114 114 Y ∗ = = ==60 ()1− a ()109−−(). 19.

Assume that Y0 = 20, then the general solution is given by,

t Yt =−( 0. 9) ( 20 − 60) + 60 =−40 − 0. 9t + 60 ()

The graph of this equation for values of t from 0 to 50 is shown below.

88 100 90 80 70 60

Y 50 40 30 20 10 0 0 1020304050 t

Here we see that Y is gradually converging on to the equilibrium value of 60 but in a cyclical manner whereby the cycles around the equilibrium are gradually diminishing. Finally, as an example of a unstable equilibrium consider the following equation:

YYtt= −11. −1 + 42

In this case a = -1.1 which is less than - 1 so the time path of Y should exhibit explosive cycles. The equilibrium value is

b 42 42 Y ∗ = = ==20 ()1− a 111−−(). 21. ()

Assuming the initial condition is Y0 = 12 the time path is shown in the following diagram.

120 100

80

60

40 Y 20

0 -20 0 5 10 15 20 25

-40 -60

t

Although an equilibrium value (of 20) exists, Y never attains this value but instead oscillates around it in ever increasing cycles. The only way equilibrium would be reached is if Y started off at the equilibrium value, i.e. if Y0 = 20.

89

5.5. Solution of a Second-Order Linear Difference Equation

Now, suppose the system had more lagged inputs, as in

yt = Ayt−1 + Byt−2 (14)

From looking at this for “most” values of A and B there is a general solution with 2 “roots”, λ1 and λ2 of the form

(15)

Now, we want to know the values of λ1and λ2. These roots are found as the solution of the “auxiliary equation”, also sometimes known as the characteristic equation.

(16)

Because we get tired of fussing over minus signs, relabel the coefficients as a1 = -_ _ A and a2 = - B.

(17)

It is “easy” to see there is at least one “test solution” like

(18)

Insert that “test solution” into (17) to get

(19)

Which can be rearranged (by dividing everything by λt−2 ) as

(20)

This is the characteristic equation for the second order dynamic system. The possible solutions are referred to as λ1 and λ2. If these two solutions are not the same, then the general solution in (15) is valid. That is the case of “distinct roots”. If the sum of the roots is less than one then the system is stable.

90 Note that for any two arbitrary constants k1 and k2 the linear combination in (15) solves the homogeneous equation.

5.6. Ordinary Differential Equations

5.6.1. A Simple Growth Model

A simple model for growth is introduced and the corresponding is derived and solved.

Example 1: The number P(t) of bacteria in a colony at time t is to be studied. The culture contains an ample food supply and there are no predators for the bacteria in the culture. The time frame under consideration is assumed to be short relative to the life expectancy of an individual bacterium. Under these conditions the only way in which the population size changes is by birth of new bacteria. One reasonable assumption is that the number of births is proportional to the number of bacteria currently alive. By equating rates of change, this assumption leads to the model

(21)

for all t > 0 where k is the proportionality constant. The constant k has the interpretation as the birth rate for the colony. The equation derived above is called an ordinary differential equation because the equation expresses a relationship between the unknown function P(t) and its derivative. (The adjective ‘ordinary’ means that only ordinary derivatives appear in the equation; there are no partial derivatives appearing in the equation.) More completely, an ordinary differential equation is an equation which expresses a relationship between the derivatives of an unknown function, the independent variable, and the unknown function itself. The equation above is a first order differential equation because only the first derivative of the unknown function P(t) appears in the equation. The equation has constant coefficients because the unknown function P(t) and its derivatives are multiplied only by constants, not by functions of the independent variable t. The equation is linear because the unknown function P(t) and its derivatives appear only to the first power. The equation is

91 also homogeneous because the function P(t ) which is identically 0 for all t satisfies the equation.

A solution of an ordinary differential equation is a function which satisfies the equation at all points in the domain of the function.

kt Exercise 1: Show that the function P(t) = e solves the differential equation above. (Note that the domain of the function ekt is all real numbers t).

d Solution: Here just substitute and check for equality. Since ekt =kekt the function ekt is a dt solution.

The previous exercise points to a two step method for solving homogeneous first order linear ordinary differential equations with constant coefficients.

mt (1) Try a solution of the form e and find the value of m that makes this function a solution. (2) The general solution is then of the form Ce mt where C is an arbitrary constant.

This procedure shows that the solution of a homogeneous first order linear ordinary differential equation with constant coefficients is an exponential function.

d Example 2: In order to solve the equation A(t) =5A(t)try a solution of the form emt . dt Substituting this trial solution into the equation leads to memt = 5emt , and this equation holds for all t only if m = 5. The general solution of the equation is A(t) = Ce5t .

Example 3: In the bacteria problem suppose the initial population is 100 and the growth rate 0.05t is 5%. Then P(t) = Ce is the general solution to the differential equation. Since the initial population is 100, P(0 ) = 100. Using this fact and the general solution shows that C = 100. Hence P(t) = 100e0.05t .

92 In the case of the previous example the given information P(0 ) = 100 is called an initial condition, for obvious reasons. The initial condition is used to find the appropriate value of the arbitrary constant that appears in the general solution of the differential equation.

5.6.2. First-Order Differential Equations

First order differential equations can always be solved (in principle) by the separation of variables technique (the equation in example 1 can also be solved using this method):

(1) Isolate the derivative of the unknown function on one side of the equation. (2) Divide by the quantity on the other side of the equation. (3) Integrate both sides of the resulting equation making use of the Fundamental Theorem of Calculus and the initial condition.

d Example 4: To illustrate the method the equation v(t) = v(t) − 1 with initial condition dt v(0) = 0 will be solved. Dividing both sides by v(t) − 1 gives

(22)

and both sides can now be integrated with respect to t. Since the initial condition is specified at time t = 0, the integration is from t = 0 to an arbitrary point t = s. This gives

(23)

After integration, using the fact that v(0) = 0, this becomes ln |v(s) − 1| = s or |v(s) − 1| = es . Since v(0) = 0, v(s) − 1 is at least initially negative, so removing the absolute values finally gives v(s) = 1 − e s for s 0 as the solution.

The success of the separation of variables technique depends on the ability to integrate the resulting expression. This can at times be difficult or even impossible, so we may need to employ computational techniques.

93 5.6.3. Second-Order Differential Equations

The general method of solving a homogeneous second order linear equation with constant coefficients is presented.

Example 5: A spring has one end attached to a vertical wall and the other end attached to a 1 kilogram mass. The mass lies on a horizontal surface and x(t) denotes the displacement from equilibrium at time t. At time 0, the displacement is 1 unit and the mass is released. According to Hooke’s Law, the force on the mass exerted by the spring is proportional to the displacement from equilibrium. The proportionality constant is called the spring constant which measures the stiffness of the spring.

Assume the spring constant is 3. Assume the frictional force is zero and the air resistance is equal in magnitude to four times the . Using Newton’s Law then gives the equation

(24)

as the equation of motion for t > 0.

The equation derived in the example is a second order linear homogeneous equation with constant coefficients. In order to solve the equation the technique used earlier in the first order case is adapted, as follows. Try a solution of the form x(t)=emt , where m is to be determined. Plugging this trial solution into the equation gives

(25)

2 and this equation will hold for all t > 0 only if m = −4m − 3. The equation m must satisfy,

2 here m + 4m + 3 = 0, is called the characteristic equation. The values of m that satisfy the characteristic equation in this case are m = −3 and m = −1. The corresponding trial solutions are e−3t and e−t The superposition principle then states that the general solution is

−3t −t x(t) = C1e + C2e . Here C1 and C2 are two arbitrary constants which are determined from the initial conditions.

94 −3t −t Exercise 2: Verify that C1e + C2e and solves the equation no matter what values C1 and

C2 have.

Solution 2: In this case the initial displacement is 1 and the initial velocity may be assumed d to be 0. This means that x(0) = 1 and x(t) = 0. Using these conditions and the general dt t=0 form of the solution gives

C + C = 1 1 2 − 3C1 − C2 = 0

1 −3t 3 −t from which C = −1/2 and C = 3/ 2. Hence x(t) = − e + e is the displacement at 1 2 2 2 time t.

The method for solving second order linear homogeneous equations with constant coefficients is therefore as follows.

(1) Try a solution of the form emt . This produces the characteristic equation (a quadratic equation) to be solved for m.

(2) Solve the characteristic equation.

(3) The general solution of the differential equation is obtained from the roots of the characteristic equation as follows.

The alternate form in the conjugate complex roots case is referred to as the phase-amplitude form. This form is useful in certain cases.

95

5.7. Introduction to Partial Differential Equations

A partial differential equation is an equation which contains partial derivatives, such as the equation ∂u ∂ 2u = (26) ∂t ∂x 2 in which u is regarded as a function of x and t .Unlike the theory of ordinary differential equations (which centers upon one key theorem, the fundamental existence and uniqueness theorem), there is no real unified theory of partial differential equations. Instead, each type of partial differential equations exhibits its own special idiosyncracies, which usually mirror the physical phenomena which it was first designed to model.

Many of the foundational theories of physics and engineering are expressed by means of systems of partial differential equations. The reader may have heard some of these equations mentioned in physics. Fluid mechanics is often formulated by the Euler equations of motion, electricity and magnetism by Maxwell equations, etc. It is therefore important to develop techniques that can be used to solve a wide variety of partial differential equations.

We will give one important simple example of partial differential equations, the heat equation, which governs propagation of heat in a bar of length L .We imagine that the bar is located along the x -axis and we let

u(x, t) = temperature at point x at time t

Heat in a small segment of the bar is proportional to temperature, the constant of proportionality being determined by the density and specific heat of the material making up the bar. If σ(x )denotes the specific heat at the point x and ρ(x ) is the density of the bar at x

,then the heat within the region D between x and x is given by the formula x 1 , x 2 1 2

x2 Heat within D = ρ(x)σ (x)u(x,t)dx (27) x1 ,x2 ∫ x1

96 Assuming that no heat is being created or destroyed in D , the rate at which heat leaves x 1 , x 2 D is: x 1 , x 2

d ⎡x2 ⎤ d ⎡x2 ∂u ⎤ − ⎢ ρ(x)σ (x)u(x,t)dx⎥ = − ⎢ ρ(x)σ (x) (x,t)dx⎥ (28) dt ∫ dt ∫ ∂t ⎣⎢ x1 ⎦⎥ ⎣⎢ x1 ⎦⎥ On the other hand, the rate of heat flow F(x,t) is proportional to the partial derivative of temperature,

∂u F(x,t) = −κ (x) (x,t) (29) ∂x where κ (x) is the thermal conductivity of the bar at x . Thus we find that the rate at which heat leaves the region D is also given by the formula x 1 , x 2

x2 ∂F F(x ,t) − F(x ,t) = (x,t)dx (30) 2 1 ∫ ∂x x1 Comparing the two formulae (28) and (30) we find that

x2 ∂F x2 ∂u (x,t)dx = − ρ(x)σ (x) (x,t)dx (31) ∫ ∂x ∫ ∂t x1 x1

This equation is true for all choices of x1 and x2 , so the integrands on the two sides must be equal: ∂F ∂u = − ρ(x)σ (x) (32) ∂x ∂t

It follows from (29) that

∂ ⎛ ∂u ⎞ ∂u ⎜−κ (x) ⎟ = − ρ(x)σ (x) (33) ∂x ⎝ ∂x ⎠ ∂t

In this way we obtain the heat equation

∂u ∂ ⎛ ∂u ⎞ ρ(x)σ (x) = ⎜κ (x) ⎟ (34) ∂t ∂x ⎝ ∂x ⎠ In the special case where the bar is homogeneous, i.e. its properties are the same at every point, ρ(x), σ (x), κ (x) are constants, say σ and κ respectively, (33) becomes

97 ∂u κ ∂ 2u = (35) ∂t ρσ ∂x 2

This is our simplest example of a linear partial differential equation. Although its most basic application concerns diffusion of heat, it arises in many other contexts as well. For example, a slight modification of the heat equation was used by Black and Scholes to price derivatives in financial markets.

98 References

1- "Fundamental Methods of Mathematical Economics", Chiang, Alpha. C. 2- "Calculus", Stewart, James, 2003. Thomson, 3- "Elementary Linear Algebra", Anton, Howard. John Wiley and Sons. 4- "Partial Differential Equations for Scientist and Engineers", Farlow, S. J. John Wiley and Sons. 5- "Econometric Analysis" Greene, William, Prentice Hall.

99

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